Let α ≠ β satisfy α2+1=6 α, β2+1=6 β. Then, the quadratic equation whose roots are (α/α+1), (β/β+1), is

Question

Let α ≠ β satisfy α2+1=6 α, β2+1=6 β. Then, the quadratic equation whose roots are (α/α+1), (β/β+1), is (A) 8 x2+8 x+1=0 (B) 8 x2-8 x-1=0 (C) 8 x2-8 x+1=0 (D) 8 x2+8 x-1=0

Tardigrade Admin · Accepted Answer

We have, α2+1=6 α and β2+1=6 β Since, α, β are the roots of the equation x2-6 x+1=0 ∴ x=(α/α+1) ⇒ α=(X/1-X) Hence, required quadratic equation whose roots are (α/α+1), (β/β+1), is ((x/1-x))2-6((x/1-x))+1=0 ⇒ x2-6 x(1-x)+1(1-x)2=0 ⇒ x2-6 x+6 x2+1+x2-2 x=0 ⇒ 8 x2-8 x+1=0

Q. Let $α \neq = β$ satisfy $α^{2} + 1 = 6 α, β^{2} + 1 = 6 β$ .
Then, the quadratic equation whose roots are $\frac{α}{α + 1}, \frac{β}{β + 1}$ , is

$8 x^{2} + 8 x + 1 = 0$

$8 x^{2} - 8 x - 1 = 0$

$8 x^{2} - 8 x + 1 = 0$

$8 x^{2} + 8 x - 1 = 0$

Q. Let α=β satisfy α2+1=6α,β2+1=6β. Then, the quadratic equation whose roots are α+1α​,β+1β​, is

8x2+8x+1=0

8x2−8x−1=0

8x2−8x+1=0

8x2+8x−1=0

We have, α2+1=6α and β2+1=6β Since, α,β are the roots of the equation x2−6x+1=0 ∴x=α+1α​ ⇒α=1−XX​ Hence, required quadratic equation whose roots are α+1α​,β+1β​, is (1−xx​)2−6(1−xx​)+1=0 ⇒x2−6x(1−x)+1(1−x)2=0 ⇒x2−6x+6x2+1+x2−2x=0 ⇒8x2−8x+1=0

Q. Let $α \neq = β$ satisfy $α^{2} + 1 = 6 α, β^{2} + 1 = 6 β$ .
Then, the quadratic equation whose roots are $\frac{α}{α + 1}, \frac{β}{β + 1}$ , is

$8 x^{2} + 8 x + 1 = 0$

$8 x^{2} - 8 x - 1 = 0$

$8 x^{2} - 8 x + 1 = 0$

$8 x^{2} + 8 x - 1 = 0$