Question

# Let a, b ∈ R,a≠0, such that the equation, ax2−2bx+5=0 has a repeated root α, which is also a root of the equation x2−2bx−10=0. If β is the other root of this equation, then α2+β2 is equal to: 24252628

Solution

## The correct option is B 25ax2−2bx+5=0 has both roots as α ⇒ 2α=2ba⇒α=ba And α2=5a ⇒b2=5a(a≠0)⋯(1)           ⇒ α+β=2b and αβ=−10 β,α are the roots of x2−2bx−10=0. α=ba ⇒b2−2ab2−10a2=0 ⇒5a−10a2−10a2=0      ( Using (1)) ⇒a=14⇒b2=54 ⇒α2=20,β2=5⇒α2+β2=25

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