If x2−3x+2 is a factor of x4−ax2+b then the equation whose roots are a,b is
A
x2−9x−20=0
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B
x2−9x+20=0
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C
x2+9x+20=0
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D
x2+9x−20=0
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Solution
The correct option is Bx2−9x+20=0 x2−3x+2=(x−1)(x−2) is factor of f(x)=x4−ax2+b
So,f(1)=0andf(2)=0 ⇒1−a+b=0and16−4a+b=0 ⇒a=5,b=4 ∴Required equation: x2−(5+4)x+5.4=0 ⇒x2−9x+20=0