Question
If α and β be the roots of the equation 2x2+2(a+b)x+a2+b2=0, then the equation whose roots are (α+β)2 and (α−β)2) is
Answer: Option B
:
B
Sum of rootsα +β = -(a + b) andαβ = a2+b22
⇒(α+β)2=(a+b)2and(α−β)2=α2+β2−2αβ
=2ab−(a2+b2)=−(a−b)2
Now the required equation whose roots are
(α+β)2and(α−β)2
x2−{(α+β)2+(α−β)2}x+(α+β)2(α−β)2=0
⇒x2−{(a+b)2+(a−b)2}x+(a+b)2(a−b)2=0
⇒x2−4abx−(a2−b2)2=0
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:
B
Sum of rootsα +β = -(a + b) andαβ = a2+b22
⇒(α+β)2=(a+b)2and(α−β)2=α2+β2−2αβ
=2ab−(a2+b2)=−(a−b)2
Now the required equation whose roots are
(α+β)2and(α−β)2
x2−{(α+β)2+(α−β)2}x+(α+β)2(α−β)2=0
⇒x2−{(a+b)2+(a−b)2}x+(a+b)2(a−b)2=0
⇒x2−4abx−(a2−b2)2=0
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