Examples with solutions for Square of Difference: Transition between expressions

Exercise #1

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

Exercise #2

(4b3)(4b3) (4b-3)(4b-3)

Rewrite the above expression as an exponential summation expression:

Video Solution

Answer

(4b3)2 (4b-3)^2

16b224b+9 16b^2-24b+9

Exercise #3

Rewrite the following expression as an addition and as a multiplication:

(3xy)2 (3x-y)^2

Video Solution

Answer

9x26xy+y2 9x^2-6xy+y^2

(3xy)(3xy) (3x-y)(3x-y)

Exercise #4

(a4)(a4)=? (a-4)(a-4)=\text{?}

Video Solution

Answer

a28a+16 a^2-8a+16

Exercise #5

Declares the given expression as a sum

(7b3x)2 (7b-3x)^2

Video Solution

Answer

49b242bx+9x2 49b^2-42bx+9x^2

Exercise #6

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

Video Solution

Answer

x=12 x=12

Exercise #7

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

Video Solution

Answer

x=3 x=3

Exercise #8

492×21+9=? 49-2\times21+9=\text{?}

Video Solution

Answer

(73)2 (7-3)^2

Exercise #9

m2943mn+4n2=? \frac{m^2}{9}-\frac{4}{3}mn+4n^2=\text{?}

Video Solution

Answer

(m32n)2 (\frac{m}{3}-2n)^2

Exercise #10

Rewrite the expression below as an addition and as a multiplication:

(mnl)2 (mn-l)^2

Video Solution

Answer

m2n22mnl+l2 m^2n^2-2mnl+l^2

(mnl)(mnl) (mn-l)(mn-l)