# Exercise 6.3

On the following page we have given the solution of Exercise 6.3 of Mathematics 9 (Science) published by Caravan Book House, Lahore.

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### Question 1:

Use factorization to find the square root of the following expressions.

(i) $4x^2-12xy +9y^2$

(ii) $x^2-1+\frac{1}{4x^2}, (x\neq 0)$

(iii) $\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2$

(iv) $4(a+b)^2-12(a^2-b^2)+9(a-b)^2$

(v) $\frac{4x^6-12x^3y^3+9y^6}{9x^4-24x^2y^2+16y^4},(x \neq 0)$

(vi) $\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right)$

(vii) $\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12,(x \neq 0)$

(viii) $(x^2+3x+2)(x^2+4x+3)(x^2+5x+6)$

(ix) $(x^2+8x+7)(2x^2-x-3)(2X^2+11x-21)$

(i) $4x^2-12xy +9y^2$

Solution:

\begin{align}4x^2-12xy +9y^2\\&=4x^2-6xy-6xy +9y^2\\&= 2x(2x-3y)-3y(2x-3y)\\&= (2x-3y)(2x-3y)\\&= (2x-3y)^2 \end{align}

\begin{align} \sqrt{4x^2-12xy +9y^2}&= \pm (2x-3y) \end{align}

(ii) $x^2-1+\frac{1}{4x^2}, (x\neq 0)$

Solution:

\begin{align}x^2-1+\frac{1}{4x^2}\\&=(x)^2-1+(\frac{1}{2x})^2\\&=(x)^2-2x(\frac{1}{2x})+(\frac{1}{2x})^2\\&= (x-\frac{1}{2x})^2 \end{align}

\begin{align} \sqrt{x^2-1+\frac{1}{4x^2}}&= \pm (x-\frac{1}{2x}) \end{align}

(iii) $\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2$

Solution:

\begin{align}\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36y^2}\\&=(\frac{1}{4}x)^2-\frac{1}{12}xy+ (\frac{1}{6y})^2\\&=(\frac{1}{4}x)^2-2(\frac{1}{4})x(\frac{1}{6y})+ (\frac{1}{6y})^2\\&= (\frac{x}{4}-\frac{1}{6y})^2 \end{align}

\begin{align} \sqrt{\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2}&= \pm (\frac{x}{4}-\frac{1}{6y}) \end{align}

iv) $4(a+b)^2-12(a^2-b^2)+9(a-b)^2$

Solution:

\begin{align}4(a+b)^2-12(a^2-b^2)+9(a-b)^2\\&=[2(a+b)]^2-2(2)(a+b)(3)(a-b)+[3(a-b)]^2\\&=(2(a+b)-3(a-b))^2\\&= (2a+2b-3a+3b)^2 \\&=(5b-a)^2 \end{align}

\begin{align} \sqrt{4(a+b)^2-12(a^2-b^2)+9(a-b)^2}&= \pm (5b-a) \end{align}

(v) $\frac{4x^6-12x^3y^3+9y^6}{9x^4+24x^2y^2+16y^4},(x \neq 0)$

Solution:

\begin{align}\frac{4x^6-12x^3y^3+9y^6}{9x^4+24x^2y^2+16y^4}\\&=\frac{(2x^3)^2-2(2x^3)(3y^3)+(3y^3)^2}{(3x^2)^2+2(3x^2)(3y^2)+(4y^2)^2}\\&=\frac{(2x^3-3y^3)^2}{(3x^2+4y^2)^2} \end{align}

\begin{align} \sqrt{\frac{4x^6-12x^3y^3+9y^6}{9x^4+24x^2y^2+16y^4}}&= \pm \frac{(2x^3-3y^3)}{(3x^2+4y^2)} \end{align}

(vi) $\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right)$

Solution:

Let $x-\frac{1}{x} =a$

Now \begin{align}\left(x+\frac{1}{x}\right)^2 &=(x-\frac{1}{x})^2+4\\&= a^2+4 \end{align}

\begin{align}\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right) &=(a^2+4)-4a \\&= a^2-4a+4\\&=(a-2)^2\\&=\left[ (x-\frac{1}{x})-2\right]^2 \end{align}

\begin{align} \sqrt{\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right)}&= \pm \left[ (x-\frac{1}{x})-2\right]\end{align}

(vii) $\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12,(x \neq 0)$

Solution:

Let $x+\frac{1}{x} =a$

Now \begin{align}\left(x+\frac{1}{x}\right)^2 &=a^2\end{align}

\begin{align}\left(x^2+\frac{1}{x^2}+2\right) &= a^2\end{align}

\begin{align}\left(x^2+\frac{1}{x^2}\right) &= a^2-2\end{align}

\begin{align}\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12 &=(a^2-2)^2-4a^2+12 \\&= a^2-4a+4\\&=a^4-4a^2+4-4a^2+12\\&=(a^2)^2-2(a^2)(4)+(4)^2\\&=(a^2-4)^2\end{align}

\begin{align} \sqrt{\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12}&= \pm \left[ (a^2-4)\right]\\&= \pm \left[(x+\frac{1}{x})^2-4 \right]\end{align}

(viii) $(x^2+3x+2)(x^2+4x+3)(x^2+5x+6)$

Solution:

\begin{align}(x^2+3x+2)(x^2+4x+3)(x^2+5x+6) &= (x^2+2x+x+2)(x^2+x+3x+3)(x^2+2x+3x+6)\\&= (x(x+2)+1(x+2))(x(x+1)+3(x+1))(x(x+2)+3(x+2))\\&= (x+2)(x+1)(x+1)(x+3)(x+2)(x+3)\\&= (x+2)^2(x+1)^2(x+3)^2\end{align}

\begin{align} \sqrt{(x^2+3x+2)(x^2+4x+3)(x^2+5x+6)}&= \pm (x+2)(x+1)(x+3)\end{align}

(ix) $(x^2+8x+7)(2x^2-x-3)(2X^2+11x-21)$

Solution:

\begin{align}(x^2+8x+7)(2x^2-x-3)(2X^2+11x-21)&= (x^2+x+7x+7)(2x^2-3x+2x-3)(2x^2+14x-3x-21)\\&=(x(x+1)+7(x+1))(x(2x-3)+1(2x-3))(2x(x+7)-3(x+7))\\&=(x+1)(x+7)(2x-3)(x+1)(x+7)(2x-3)\\&= (x+1)^2(x+7)^2(2x-3)^2\end{align}

\begin{align} \sqrt{(x^2+8x+7)(2x^2-x-3)(2X^2+11x-21)}&= \pm (x+1)^2(x+7)^2(2x-3)^2\end{align}

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