Question

find the value of x 9x^2-9(p+q)x+(2p^2+5pq+2q^2)=0

Answer 1

x=(p+2q)/3
x=(2p+q)/3

Answer 2

Answer:

$value \:of \: x = \frac{2p+q}{3}\:Or\:\frac{p+2q}{3}$

Step-by-step explanation:

Given Quadratic equation,

9x²-9(p+q)x+(2p²+5pq+2q²)=0,

Compare the equation ,

with ax²+bx+c=0, we get

a= 9, b = -9(p+q),

c = 2p²+5pq+2q²

= 2(p²+2pq+q²)+pq

= 2(p+q)²+pq,

Now,

Discreminant (D)=b²-4ac

= [-9(p+q)]²-4×9×[2(p+q)²+pq]

= 81(p+q)²-36[2(p+q)²+pq]

= 81(p+q)²-72(p+q)²-36pq

= 9(p+q)²-36pq

= 9[(p+q)²-4pq]

=9[p²+q²+2pq-4pq]

= 9(p²+q²-2pq)

=9(p-q)²

D=[3(p-q)]²

Now,

$x = \frac{-b±\sqrt{D}}{2a}$

$\implies x = \frac{-[-(9(p+q)]±\sqrt{[3(p-q)]^{2}}}{2\times 9}$

$=\frac{9(p+q)±3(p-q)}{18}\\=\frac{3[3(p+q)±(p-q)}{18}\\=\frac{3p+3q±(p-q)}{6}$

Now,

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

$\implies x = \frac{3p+3q+p-q}{6}\:Or\:\frac{3p+3q-p+q)}{6}$

$\implies x = \frac{4p+2q}{6}\:Or\:\frac{2p+4q}{6}$

$\implies x = \frac{2p+q}{3}\:Or\:\frac{p+2q}{3}$

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