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Question

Solve the following quadratic equation using quadratic formula .

9x29(a+b)x+(2a2+5ab+2b2)=0


A

The roots are 2a+b3 and a+2b3

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B

The roots are 2a+b3 and a2b3

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C

The roots are 5a+b3 and a+2b3

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D

The roots are 2a+b3 and a2b4

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Solution

The correct option is

A

The roots are 2a+b3 and a+2b3



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) and C=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 unequal 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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