Question

Solve the following equation for x : 9x square - 9(p+q)x + (2p + 5pq + 2q square ) = 0

Answer 1

Answer:

Step-by-step explanation:mark me as brainliest plz

Answer 2

Answer:

The solution of the expression is $x=\frac{2p+q}{3},\frac{p+2q}{3}$          

Step-by-step explanation:

Given : Equation $9x^2-9(p+q)x+(2p^2+5pq+2q^2)=0$

To find : Solve the following equation for x?

Solution :

Applying quadratic formula of equation $ax^2+bx+c=0$ solution is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

On comparing, $a=9, b=-9(p+q), c=2p^2+5pq+2q^2$

First we find,

$b^2-4ac=(-9(p+q))^2-4(9)(2p^2+5pq+2q^2)$

$b^2-4ac=81(p^2+q^2+2pq)-36(2p^2+5pq+2q^2)$

$b^2-4ac=81p^2+81q^2+162pq-72p^2-180pq-72q^2$

$b^2-4ac=9p^2+9q^2-18pq$

$b^2-4ac=(3p)^2+(3q)^2-2(3p)(3q)$

$b^2-4ac=(3p-3q)^2$

Substitute in the equation,

$x=\frac{-(-9(p+q))\pm\sqrt{(3p-3q)^2}}{2(9)}$

$x=\frac{9(p+q)\pm (3p-3q)}{18}$

$x=\frac{3(p+q)\pm (p-q)}{6}$

$x=\frac{3p+3q+p-q}{6},\frac{3p+3q-p+q}{6}$

$x=\frac{4p+2q}{6},\frac{2p+4q}{6}$

$x=\frac{2p+q}{3},\frac{p+2q}{3}$

Therefore, The solution of the expression is $x=\frac{2p+q}{3},\frac{p+2q}{3}$