Solve the following quadratic equation using quadratic formula . 9x2−9(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, 9x2−9(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=B2−4AC ⇒D=81(a+b)2−36(2a2+5ab+2b2) ⇒D=81(a2+b2+2ab)−(72a2+180ab+72b2) ⇒D=9a2+9b2−18ab ⇒D=9(a2+b2−2ab) ⇒D=9(a−b)2≥0 ⇒D≥0 So, the roots of the given equation are real and are given by α=−B+√D2A=9(a+b)+3(a−b)18=12a+6b18=2a+b3 and, β=−B−√D2A=9(a+b)−3(a−b)18=6a+12b18=a+2b3
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