Question

Solve the following equations by factorisation method.Only factorisation not dharacharya.​

Answer 1

Hello, please consider the following.

When $x_1$ and $x_2$ are two roots, we can factorise as

$ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0$

So for the first equation, we can say that the sum of the zeros is

$\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}$

and the product is

$\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}$

So we can factorise as below.

$4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0$

And the solutions are

$\boxed{\sf \n\bf \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}$

For the second equation, we will complete the square and put the constant on the right side and take the root.

Let's do it!

$9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}$

We take the root, and we find the two solutions

$\begin{aligned}x_1&=\dfrac{a+b}{2}-\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b-a+b}{6}\\\\&=\dfrac{2a+4b}{6}\\\\&\boxed{=\dfrac{a+2b}{3}}\end{aligned}$

$\begin{aligned}x_1&=\dfrac{a+b}{2}+\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b+a-b}{6}\\\\&=\dfrac{4a+2b}{6}\\\\&\boxed{=\dfrac{2a+b}{3}}\end{aligned}$

Thank you.