Square of Difference: Solving the problem

Equations with denominatorsUsing quadrilateralsComplete the missing numbersData with powers and rootsMore than one factorizationNumber of termsIdentify the greater valueUsing fractionsEquations with variables on both sidesResulting in a quadratic equationSystem of equations with no solutionUsing multiple rulesTransition between expressions
Question 1

\( 60-16y+y^2=-4 \)

Question 2

\( x^2+144=24x \)

Question 3

\( x^2=6x-9 \)

Question 4

\( (x-4)^2-x(x+8)=0 \)

Question 5

\( (x-4)^2-x(x+8)=16 \)

Examples with solutions for Square of Difference: Solving the problem

Exercise #1

60−16y+y2=−4 60-16y+y^2=-4

Video Solution

Step-by-Step Solution

Let's solve the given equation:

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

60−16y+y2=−460−16y+y2+4=0y2−16y+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:

(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 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:

y2−16y+64=0↓y2−16y+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:

(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 of the equation we got in the last step:

y2−16y‾+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):

(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 it's possible to present the expression on the left side of the equation as:

y2−16y‾+82=0↕?y2−2⋅y⋅8‾+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:

2⋅y⋅8=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:

y2−2⋅y⋅8+82=0↓(y−8)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:

(y−8)2=0/y−8=±0y−8=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:

60−16y+y2=−4y2−16y+64=0↓y2−2⋅y⋅8+82=0↓(y−8)2=0↓y−8=0↓y=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

Exercise #2

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

Video Solution

Answer

x=12 x=12

Exercise #3

x2=6x−9 x^2=6x-9

Video Solution

Answer

x=3 x=3

Exercise #4

(x−4)2−x(x+8)=0 (x-4)^2-x(x+8)=0

Video Solution

Answer

x=1 x=1

Exercise #5

(x−4)2−x(x+8)=16 (x-4)^2-x(x+8)=16

Video Solution

Answer

x=0 x=0

Question 1

\( (x-1)^2-(x+2)^2=15 \)

Question 2

\( x^2+(x-2)^2=2(x+1)^2 \)

Question 3

Solve for x:

\( (x-4)^2=(x+2)(x-1) \)

Question 4

Solve the following equation:

\( \frac{1}{(x-2)^2}+\frac{1}{x-2}=1 \)

Question 5

Solve the following equation:

\( \frac{x^3+1}{(x-1)^2}=x+4 \)

Exercise #6

(x−1)2−(x+2)2=15 (x-1)^2-(x+2)^2=15

Video Solution

Answer

x=−3 x=-3

Exercise #7

x2+(x−2)2=2(x+1)2 x^2+(x-2)^2=2(x+1)^2

Video Solution

Answer

x=14 x=\frac{1}{4}

Exercise #8

Solve for x:

(x−4)2=(x+2)(x−1) (x-4)^2=(x+2)(x-1)

Video Solution

Answer

x=2 x=2

Exercise #9

Solve the following equation:

1(x−2)2+1x−2=1 \frac{1}{(x-2)^2}+\frac{1}{x-2}=1

Video Solution

Answer

12[5±5] \frac{1}{2}[5\pm\sqrt{5}]

Exercise #10

Solve the following equation:

x3+1(x−1)2=x+4 \frac{x^3+1}{(x-1)^2}=x+4

Video Solution

Answer

x=3,12 x=3,\frac{1}{2}

Question 1

\( \frac{(\frac{1}{x}-\frac{1}{2})^2}{(\frac{1}{x}-\frac{1}{3})^2}=\frac{9}{4} \)

Find X

Question 2

Solve the following system of equations:

\( \begin{cases} \sqrt{x}-\sqrt{y}=\sqrt{\sqrt{61}-6} \\ xy=9 \end{cases} \)

Question 3

Solve the following equation:

\( \frac{(2x-1)^2}{x-2}+\frac{(x-2)^2}{2x-1}=4.5x \)

Exercise #11

(1x−12)2(1x−13)2=94 \frac{(\frac{1}{x}-\frac{1}{2})^2}{(\frac{1}{x}-\frac{1}{3})^2}=\frac{9}{4}

Find X

Video Solution

Answer

2.5

Exercise #12

Solve the following system of equations:

{x−y=61−6xy=9 \begin{cases} \sqrt{x}-\sqrt{y}=\sqrt{\sqrt{61}-6} \\ xy=9 \end{cases}

Video Solution

Answer

x=612−2.5 x=\frac{\sqrt{61}}{2}-2.5

y=612+2.5 y=\frac{\sqrt{61}}{2}+2.5

or

x=612+2.5 x=\frac{\sqrt{61}}{2}+2.5

y=612−2.5 y=\frac{\sqrt{61}}{2}-2.5

Exercise #13

Solve the following equation:

(2x−1)2x−2+(x−2)22x−1=4.5x \frac{(2x-1)^2}{x-2}+\frac{(x-2)^2}{2x-1}=4.5x

Video Solution

Answer

−1±3 -1\pm\sqrt{3}