The Quadratic Formula: Applying the formula with two solutions

Let's recall the quadratic formula:

We'll substitute the given data into the formula:

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

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

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

$x={{10\pm\sqrt{196}\over 4}}$

$x={{10\pm14\over 4}}$

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

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

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

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

This is a quadratic equation:

$x^2+3x-18=0$

This is due to the fact that 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 to solve it using the quadratic formula,

Remember:

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

$ax^2+bx+c=0$

are:

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

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+3x-18=0$

And solve it:

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

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

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

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

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-18)}}{2\cdot1}$

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

$x_{1,2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}$

Therefore the solutions of the equation are:

$$$\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.

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

This is a quadratic equation:

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

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it to a form where all the terms on one side are ordered from the highest to the 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 to solve it using the quadratic formula,

Remember:

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

$ax^2+bx+c=0$

are:

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

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

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

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

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

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

We obtain the equation's solutions (roots) by inserting the coefficients we just noted into the quadratic formula:

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

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

$x_{1,2}=\frac{-5\pm\sqrt{1}}{2}=\frac{-5\pm1}{2}$

The solutions to the equation are:

$$$\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.

This is a quadratic equation:

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

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in to a form where all terms on one side are ordered from the highest to the 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 to solve it using the quadratic formula.

Remember:

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

$ax^2+bx+c=0$

are:

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

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

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

And solve it:

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

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

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

We obtain the solutions of the equation (its roots) by insertion we just identified into the quadratic formula:

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

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

$x_{1,2}=\frac{-5\pm\sqrt{9}}{2}=\frac{-5\pm3}{2}$

Therefore the solutions of the equation are:

$$$\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.

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

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

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:

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

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

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

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

$x = {-10 \pm 8 \over 2}$

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

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

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

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

And that's the solution!

In order to solve the equation, start by removing the denominators.

To do this, we'll multiply the denominators:

$(2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)$

Open the parentheses on the left side, making use of the distributive property:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)$

Continue to open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$

Simplify further:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x$

Go back and simplify the parentheses on the left side of the equation:

$8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x$

Combine like terms:

$9x^3+18x^2+18x+9=9x^3+22.5x+9x$

Notice that all terms can be divided by 9 as shown below:

$x^3+2x^2+2x+1=x^3+2.5x+x$

Move all numbers to one side:

$x^3-x^3+2x^2-2.5x^2+2x-x+9=0$

We obtain the following:

$0.5x^2-x-1=0$

In order to remove the one-half coefficient, multiply the entire equation by 2

$x^2-2x-2=0$

Apply the square root formula, as shown below-

$\frac{2±\sqrt{12}}{2}$

Apply the properties of square roots in order to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Divide both the numerator and denominator by 2 as follows:

$1±\sqrt{3}$