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Question
Solve the following quadratic equation using quadratic formula .
 9x29(a+b)x+(2a2+5ab+2b2)=0
Options:
A .  The roots are 2a+b3 and a+2b3
B .  The roots are 2a+b3 and a−2b3
C .  The roots are 5a+b3 and a+2b3
D .  The roots are 2a+b3 and a−2b4
Answer: Option A
:
A
We have,
9x29(a+b)x+(2a2+5ab+2b2)=0
Comparing this equation with Ax2+Bx+C=0, we have
A=9,B=9(a+b)andC=2a2+5ab+2b2
D=B24AC
D=81(a+b)236(2a2+5ab+2b2)
D=81(a2+b2+2ab)(72a2+180ab+72b2)
D=9a2+9b218ab
D=9(a2+b22ab)
D=9(ab)20
D0
So, the roots of the given equation are real and are given by
α=B+D2A=9(a+b)+3(ab)18=12a+6b18=2a+b3
and, β=BD2A=9(a+b)3(ab)18=6a+12b18=a+2b3

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