9x2-9(p+q)x+(2p2+5pq+2q2)solve this for x
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Sia ? 6 years, 7 months ago
Given
{tex}9 x ^ { 2 } - 9 ( a + b ) x + 2 a ^ { 2 } + 5 a b + 2 b ^ { 2 } = 0{/tex}
Now, {tex}2 a ^ { 2 } + 5 a b + 2 b ^ { 2 } = 2 a ^ { 2 } + 4 a b + a b + 2 b ^ { 2 }{/tex}
{tex}= 2 a [ a + 2 b ] + b [ a + 2 b ]{/tex}
= (a + 2b) (2a + b)
Hence the given equation becomes,
{tex}9 x ^ { 2 } - 9 ( a + b ) x + ( a + 2 b ) ( 2 a + b ) = 0{/tex}
or, {tex}9 x ^ { 2 } - 3 [ 3 a + 3 b ] x + ( a + 2 b ) ( 2 a + b ) = 0{/tex}
{tex}9x^2-3(a+2b)x-3(2a+b)x+(a+2b)(2a+b)=0{/tex}
{tex}3x[3x-(a+2b)]-(2a+b)[3x-(a+2b)]=0{/tex}
or, {tex}[ 3 x - ( a + 2 b ) ] [ 3 x - ( 2 a + b ) ] = 0{/tex}
{tex}\Rightarrow\ either\ 3x=a+2b\ or\ 3x=2a+b{/tex}
{tex}\Rightarrow\ either\ x=\frac{a+2b}{3}\ or\ x=\frac{2a+b}{3}{/tex}
Hence, the roots = {tex}\frac { a + 2 b } { 3 } , \frac { 2 a + b } { 3 }{/tex}
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