Methods for solving a quadratic function

In this article, we will learn the three most common ways to solve a quadratic function easily and quickly.

  1. Trinomial
  2. Quadratic Formula
  3. Completing the Square

Reminder:

The basic quadratic function equation is:
Y=ax2+bx+cY=ax^2+bx+c

When:
a a   - the coefficient of X2X^2
b b   - the coefficient of XX
cc - the constant term

  • aa must be different from 00
  • bb or cc can be 00
  • a,b,ca,b,c can be negative/positive
  • The quadratic function can also look like this:
    • Y=ax2Y=ax^2
    • Y=ax2+bxY=ax^2+bx
    • Y=ax2+cY=ax^2+c

Practice Solving Quadratic Equations

Examples with solutions for Solving Quadratic Equations

Exercise #1

Solve the following equation:

2x210x12=0 2x^2-10x-12=0

Video Solution

Step-by-Step Solution

Let's recall the quadratic formula:

Quadratic formula | The formula

We'll substitute the given data into the formula:

x=(10)±10242(12)22 x={{-(-10)\pm\sqrt{-10^2-4\cdot2\cdot(-12)}\over 2\cdot2}}

Let's simplify and solve the part under the square root:

x=10±100+964 x={{10\pm\sqrt{100+96}\over 4}}

x=10±1964 x={{10\pm\sqrt{196}\over 4}}

x=10±144 x={{10\pm14\over 4}}

Now we'll solve using both methods, once with the addition sign and once with the subtraction sign:

x=10+144=244=6 x={{10+14\over 4}} = {24\over4}=6

x=10144=44=1 x={{10-14\over 4}} = {-4\over4}=-1

We've arrived at the solution: X=6,-1

Answer

x1=6 x_1=6 x2=1 x_2=-1

Exercise #2

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term


What is the value of c c in the function y=x2+25x y=-x^2+25x ?

Video Solution

Step-by-Step Solution

Let's recall the general form of the quadratic function:

y=ax2+bx+c y=ax^2+bx+c The function given in the problem is:

y=x2+25x y=-x^2+25x c c is the free term (meaning the coefficient of the term with power 0),

In the function in the problem there is no free term,

Therefore, we can identify that:

c=0 c=0 Therefore, the correct answer is answer A.

Answer

c=0 c=0

Exercise #3

Solve the following equation:

x2+3x18=0 x^2+3x-18=0

Video Solution

Step-by-Step Solution

Notice that the quadratic equation:

x2+3x18=0 x^2+3x-18=0

and this is because there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of the equation of the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

x2+3x18=0 x^2+3x-18=0

And solve it:

First, let's identify the coefficients of the terms:

{a=1b=3c=18 \begin{cases}a=1\\b=3\\c=-18\end{cases}

where we noted that the coefficient of the quadratic term is 1,

And we'll get the solutions of the equation (its roots) by substituting the coefficients we just noted in the quadratic formula:

x1,2=b±b24ac2a=3±3241(18)21 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-18)}}{2\cdot1}

Let's continue and calculate the expression inside the square root and simplify the expression:

x1,2=3±812=3±92 x_{1,2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}

Therefore the solutions of the equation are:

{x1=3+92=3x2=392=6 \begin{cases}x_1=\frac{-3+9}{2}=3 \\ x_2=\frac{-3-9}{2}=-6\end{cases}

Therefore the correct answer is answer C.

Answer

x1=3,x2=6 x_1=3,x_2=-6

Exercise #4

What is the value of X in the following equation?

X2+10X+9=0 X^2+10X+9=0

Video Solution

Step-by-Step Solution

To answer the question, we'll need to recall the quadratic formula:

x=b±b24ac2a x = {-b \pm \sqrt{b^2-4ac} \over 2a}

 

Let's remember that:

a is the coefficient of X²

b is the coefficient of X

c is the free term

 

And if we look again at the formula given to us:

a=1

b=10

c=9

 

Let's substitute into the formula:

x=10±10241921 x = {-10 \pm \sqrt{10^2-4\cdot 1 \cdot 9} \over 2\cdot 1}

Let's start by solving what's under the square root:

x=10±100362 x = {-10 \pm \sqrt{100-36} \over 2}

x=10±642 x = {-10 \pm \sqrt{64} \over 2}

x=10±82 x = {-10 \pm 8 \over 2}

Now we'll solve twice, once with plus and once with minus

 

x=10+82=22=1 x = {-10 +8 \over 2}= {-2 \over 2} = -1

x=1082=182=9 x = {-10 -8 \over 2} = {-18 \over 2} =-9

And we can see that we got two solutions, X=-1 and X=-9

And that's the solution!

Answer

x1=1,x2=9 x_1=-1,x_2=-9

Exercise #5

Solve the following equation:

x2+5x+6=0 x^2+5x+6=0

Video Solution

Step-by-Step Solution

Notice that the quadratic equation:

x2+5x+6=0 x^2+5x+6=0

and this is because there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of an equation of the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

x2+5x+6=0 x^2+5x+6=0 and solve it:

First, let's identify the coefficients of the terms:

{a=1b=5c=6 \begin{cases}a=1 \\ b=5 \\ c=6\end{cases}

where we noted that the coefficient of the quadratic term is 1,

And we'll get the equation's solutions (roots) by substituting the coefficients we just noted into the quadratic formula:

x1,2=b±b24ac2a=5±5241621 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}

Let's continue and calculate the expression inside the square root and simplify the expression:

x1,2=5±12=5±12 x_{1,2}=\frac{-5\pm\sqrt{1}}{2}=\frac{-5\pm1}{2}

Therefore the solutions to the equation are:

{x1=5+12=2x2=512=3 \begin{cases}x_1=\frac{-5+1}{2}=-2 \\ x_2=\frac{-5-1}{2}=-3\end{cases}

Therefore the correct answer is answer D.

Answer

x1=3,x2=2 x_1=-3,x_2=-2

Exercise #6

Solve the following equation:

x2+5x+4=0 x^2+5x+4=0

Video Solution

Step-by-Step Solution

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

 

We substitute into the formula:

 

-5±√(5²-4*1*4) 
          2

 

-5±√(25-16)
         2

 

-5±√9
    2

 

-5±3
   2

 

The symbol ± means that we have to solve this part twice, once with a plus and a second time with a minus,

This is how we later get two results.

 

-5-3 = -8
-8/2 = -4

 

-5+3 = -2
-2/2 = -1

 

And thus we find out that X = -1, -4

Answer

x1=1 x_1=-1 x2=4 x_2=-4

Exercise #7

Solve the following:

x2+5x+4=0 x^2+5x+4=0

Video Solution

Step-by-Step Solution

Notice that the quadratic equation:

x2+5x+4=0=0 x^2+5x+4=0 =0

and this is because there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of an equation of the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

x2+5x+4=0=0 x^2+5x+4=0 =0

And solve it:

First, let's identify the coefficients of the terms:

{a=1b=5c=4 \begin{cases}a=1 \\ b=5 \\ c=4\end{cases}

where we noted that the coefficient of the quadratic term is 1,

And we'll get the solutions of the equation (its roots) by substituting the coefficients we just identified into the quadratic formula:

x1,2=b±b24ac2a=5±5241421 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot4}}{2\cdot1}

Let's continue and calculate the expression inside the square root and simplify the expression:

x1,2=5±92=5±32 x_{1,2}=\frac{-5\pm\sqrt{9}}{2}=\frac{-5\pm3}{2}

Therefore the solutions of the equation are:

{x1=5+32=1x2=532=4 \begin{cases}x_1=\frac{-5+3}{2}=-1 \\ x_2=\frac{-5-3}{2}=-4\end{cases}

Therefore the correct answer is answer C.

Answer

x1=1,x2=4 x_1=-1,x_2=-4

Exercise #8

x2+9=0 x^2+9=0

Solve the equation

Video Solution

Step-by-Step Solution

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

 

We identify that we have:
a=1
b=0
c=9

 

We recall the root formula:

Roots formula | The version

We replace according to the formula:

-0 ± √(0²-4*1*9)

           2

 

We will focus on the part inside the square root (also called delta)

√(0-4*1*9)

√(0-36)

√-36

 

It is not possible to take the square root of a negative number.

And so the question has no solution.

Answer

No solution

Exercise #9

x2+10x=25 x^2+10x=-25

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x2+10x=25 x^2+10x=-25

First, let's arrange the equation by moving terms:

x2+10x=25x2+10x+25=0 x^2+10x=-25 \\ x^2+10x+25=0 \\ Now we notice that the expression on the left side can be factored using the perfect square trinomial formula for a binomial squared:

(a+b)2=a2+2ab+b2 (\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2

We can do this using the fact that:

25=52 25=5^2

Therefore, we'll represent the rightmost term as a squared term:

x2+10x+25=0x2+10x+52=0 x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0

Now let's examine again the perfect square trinomial formula mentioned earlier:

(a+b)2=a2+2ab+b2 (\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

And the expression on the left side in the equation we got in the last step:

x2+10x+52=0 \textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0

Notice that the terms x2,52 \textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a line):

(a+b)2=a2+2ab+b2 (\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

In other words - we'll ask if we can represent the expression on the left side as:

x2+10x+52=0?x2+2x5+52=0 \textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0

And indeed it is true that:

2x5=10x 2\cdot x\cdot5=10x

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

x2+2x5+52=0(x+5)2=0 \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0

From here we can take the square root of both sides of the equation (and don't forget there are two possibilities - positive and negative when taking the square root of an even power), then we'll easily solve by isolating the variable:

(x+5)2=0/x+5=±0x+5=0x=5 (x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}

Let's summarize the solution of the equation:

x2+10x=25x2+10x+25=0x2+2x5+52=0(x+5)2=0x+5=0x=5 x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}

Therefore the correct answer is answer C.

Answer

x=5 x=-5

Exercise #10

Solve the following equation:

2x23x+5=0 2x^2-3x+5=0

Video Solution

Step-by-Step Solution

Let's identify that this is a quadratic equation:

2x23x+5=0 2x^2-3x+5=0

and this is because there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of an equation in the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

2x23x+5=0 2x^2-3x+5=0 and solve it:

First, let's identify the coefficients of the terms:

{a=2b=3c=5 \begin{cases}a=2 \\ b=-3 \\ c=5\end{cases}

where we noted that the coefficient includes the minus sign, and this is because in the general form of the equation we mentioned earlier:

ax2+bx+c=0 ax^2+bx+c=0

the coefficients are defined such that they have a plus sign in front of them, and therefore the minus sign must be included in the coefficient value.

Let's continue and get the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:

x1,2=b±b24ac2a=(3)±(3)242522 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot2\cdot5}}{2\cdot2}

Let's continue and calculate the expression under the root and simplify the expression:

x1,2=3±312 x_{1,2}=\frac{3\pm\sqrt{-31}}{2}\frac{}{}

We got that the expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of x x that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

Answer

No solution

Exercise #11

Solve the following equation:

x2+6x10=0 -x^2+6x-10=0

Video Solution

Step-by-Step Solution

First, we'll identify that this is a quadratic equation:

x2+6x10=0 -x^2+6x-10=0

and this is because there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

For ease of solving and minimizing errors, it is always recommended to ensure that the coefficient of the quadratic term in the equation is positive,

We'll achieve this by multiplying (both sides of) the equation by:1 -1 :

x2+6x10=0/(1)x26x+10=0 -x^2+6x-10=0 \hspace{8pt}\text{/}\cdot(-1)\\ x^2-6x+10=0

Let's continue solving the equation:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of an equation of the form:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

This formula is called: "The Quadratic Formula"

Let's return to the problem:

x26x+10=0 x^2-6x+10=0

and solve it:

First, let's identify the coefficients of the terms:

{a=1b=6c=10 \begin{cases}a=1 \\ b=-6 \\ c=10\end{cases}

where we noted that the coefficient of the quadratic term is 1,

And we'll get the equation's solutions (roots) by substituting these coefficients that we mentioned earlier in the quadratic formula:

x1,2=b±b24ac2a=(6)±(6)2411021 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot10}}{2\cdot1}

Let's continue and calculate the expression under the root and simplify the expression:

x1,2=6±42 x_{1,2}=\frac{6\pm\sqrt{-4}}{2}

We got that the expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of x x that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

Answer

No solution

Exercise #12

16a2+20a+20=520a 16a^2+20a+20=-5-20a

Video Solution

Step-by-Step Solution

Let's solve the given equation:

16a2+20a+20=520a 16a^2+20a+20=-5-20a

First, let's organize the equation by moving terms and combining like terms:

16a2+20a+20=520a16a2+20a+20+5+20a=016a2+40a+25=0 16a^2+20a+20=-5-20a \\ 16a^2+20a+20+5+20a =0\\ 16a^2+40a+25=0

Now let's note that we can factor the expression on the left side using the perfect square trinomial formula for a binomial squared:

(x+y)2=x2+2xy+y2 (\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+2\textcolor{red}{x}\textcolor{blue}{y}+\textcolor{blue}{y}^2

We'll do this using the fact that:

16=4225=52 16=4^2\\ 25=5^2

And using the law of exponents for powers applied to products in parentheses (in reverse):

xnyn=(xy)n x^ny^n=(xy)^n

Therefore, first we'll express the outer terms as a product of squared terms:

16a2+40a+25=042a2+40a+52=0(4a)2+40a+52=0 16a^2+40a+25=0 \\ 4^2a^2+40a+5^2=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+40a+\textcolor{blue}{5}^2=0

Now let's examine again the perfect square trinomial formula mentioned earlier:

(x+y)2=x2+2xy+y2 (\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2

And the expression on the left side of the equation that we got in the last step:

(4a)2+40a+52=0 (\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0

Note that the terms (4a)2,52 (\textcolor{red}{4a})^2,\hspace{6pt}\textcolor{blue}{5}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

(x+y)2=x2+2xy+y2 (\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2

In other words - we ask if we can express the expression on the left side of the equation as:

(4a)2+40a+52=0?(4a)2+24a5+52=0 (\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ (\textcolor{red}{4a})^2+\underline{2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0

And indeed it holds that:

24a5=40a 2\cdot4a\cdot5=40a

Therefore, we can express the expression on the left side of the equation as a perfect square binomial:

(4a)2+24a5+52=0(4a+5)2=0 (\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable and dividing both sides of the equation by the variable's coefficient:

(4a+5)2=0/4a+5=±04a+5=04a=5/:4a=54 (4a+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ 4a+5=\pm0\\ 4a+5=0\\ 4a=-5\hspace{8pt}\text{/}:4\\ \boxed{a=-\frac{5}{4}}

Let's summarize the solution of the equation:

16a2+20a+20=520a16a2+40a+25=0(4a)2+24a5+52=0(4a+5)2=04a+5=0a=54 16a^2+20a+20=-5-20a \\ 16a^2+40a+25=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ 4a+5=0\\ \downarrow\\ \boxed{a=-\frac{5}{4}}

Therefore the correct answer is answer D.

Answer

x=54 x=-\frac{5}{4}

Exercise #13

x2+144=24x x^2+144=24x

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x2+144=24x x^2+144=24x

First, let's arrange the equation by moving terms:

x2+144=24xx224x+144=0 x^2+144=24x \\ x^2-24x+144=0

Now let's notice that we can factor the expression on the left side using the perfect square trinomial formula:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2

We can do this using the fact that:

144=122 144=12^2

Therefore, we'll represent the rightmost term as a squared term:

x224x+144=0x224x+122=0 x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{12}^2=0

Now let's examine again the perfect square trinomial formula mentioned earlier:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

And the expression on the left side in the equation we got in the last step:

x224x+122=0 \textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0

Notice that the terms x2,122 \textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{12}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

In other words - we'll ask if we can represent the expression on the left side of the equation as:

x224x+122=0?x22x12+122=0 \textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0

And indeed it is true that:

2x12=24x 2\cdot x\cdot12=24x

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

x22x12+122=0(x12)2=0 \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

(x12)2=0/x12=±0x12=0x=12 (x-12)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-12=\pm0\\ x-12=0\\ \boxed{x=12}

Let's summarize the solution of the equation:

x2+144=24xx224x+144=0x22x12+122=0(x12)2=0x12=0x=12 x^2+144=24x \\ x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0 \\ \downarrow\\ x-12=0\\ \downarrow\\ \boxed{x=12}

Therefore the correct answer is answer C.

Answer

x=12 x=12

Exercise #14

x2=6x9 x^2=6x-9

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x2=6x9 x^2=6x-9

First, let's arrange the equation by moving terms:

x2=6x9x26x+9=0 x^2=6x-9 \\ x^2-6x+9=0

Now let's notice that we can factor the expression on the left side using the perfect square trinomial formula for a binomial squared:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2

We'll do this using the fact that:

9=32 9=3^2 Therefore, we'll represent the rightmost term as a squared term:

x26x+9=0x26x+32=0 x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-6x+\textcolor{blue}{3}^2=0

Now let's examine again the perfect square trinomial formula mentioned earlier:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

And the expression on the left side in the equation we got in the last step:

x26x+32=0 \textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0

Notice that the terms x2,32 \textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{3}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

In other words - we'll ask if we can represent the expression on the left side of the equation as:

x26x+32=0?x22x3+32=0 \textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0

And indeed it is true that:

2x3=6x 2\cdot x\cdot3=6x

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

x22x3+32=0(x3)2=0 \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

(x3)2=0/x3=±0x3=0x=3 (x-3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-3=\pm0\\ x-3=0\\ \boxed{x=3}

Let's summarize the solution of the equation:

x2=6x9x26x+9=0x22x3+32=0(x3)2=0x3=0x=3 x^2=6x-9 \\ x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0 \\ \downarrow\\ x-3=0\\ \downarrow\\ \boxed{x=3}

Therefore the correct answer is answer C.

Answer

x=3 x=3

Exercise #15

6016y+y2=4 60-16y+y^2=-4

Video Solution

Step-by-Step Solution

Let's solve the given equation:

6016y+y2=4 60-16y+y^2=-4 First, let's arrange the equation by moving terms:

6016y+y2=46016y+y2+4=0y216y+64=0 60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0 Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2 This is done using the fact that:

64=82 64=8^2 So let's present the outer term on the right as a square:

y216y+64=0y216y+82=0 y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0 Now let's examine again the short factoring formula we mentioned earlier:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 And the expression on the left side of the equation we got in the last step:

y216y+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0 Let's note that the terms y2,82 \textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2 indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, meaning the middle term in the expression (underlined):

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

y216y+82=0?y22y8+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0 And indeed it holds that:

2y8=16y 2\cdot y\cdot8=16y So we can present the expression on the left side of the given equation as a difference of two squares:

y22y8+82=0(y8)2=0 \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

(y8)2=0/y8=±0y8=0y=8 (y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y-8=\pm0\\ y-8=0\\ \boxed{y=8}

Let's summarize then the solution of the equation:

6016y+y2=4y216y+64=0y22y8+82=0(y8)2=0y8=0y=8 60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}

So the correct answer is answer a.

Answer

y=8 y=8