Question
Solve the following quadratic equation using quadratic formula .
9x2−9(a+b)x+(2a2+5ab+2b2)=0
9x2−9(a+b)x+(2a2+5ab+2b2)=0
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
Was this answer helpful ?
:
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
Was this answer helpful ?
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