Factorise: frac{25}{4} x^{2}-frac{y^{2}}{9} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We have the expression: $\frac{25}{4} x^{2}-\frac{y^{2}}{9}$This can be written as: $\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}$U

Vedclass · Accepted Answer

$\left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)$

Vedclass · Answer

(A) We have the expression: $\frac{25}{4} x^{2}-\frac{y^{2}}{9}$This can be written as: $\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}$Using the algebraic identity $a^{2}-b^{2} = (a+b)(a-b)$,where $a = \frac{5}{2} x$ and $b = \frac{y}{3}$,we get:$\frac{25}{4} x^{2}-\frac{y^{2}}{9} = \left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)$

Factorise: $\frac{25}{4} x^{2}-\frac{y^{2}}{9}$

Similar Questions

Expand each of the following,using suitable identities: $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$

Find the value of $k$, if $x - 1$ is a factor of $p(x)$ in this case: $p(x) = kx^2 - \sqrt{2}x + 1$.

Factorise: $2 x^{2}+7 x+3$

Determine which of the following polynomials has $(x + 1)$ as a factor: $x^{3} + x^{2} + x + 1$.

Factorise: $3x^{2}-x-4$

Vedclass Test Series

Exam Paper Generator

Online Exam Module