Let a, b ∈ R be such that the equation a x2-2 b x+15=0 has a repeated root α. If α and β are the roots of the equation x2-2 b x+21=0, then α2+β2 is equal to:

Question

Let a, b ∈ R be such that the equation a x2-2 b x+15=0 has a repeated root α. If α and β are the roots of the equation x2-2 b x+21=0, then α2+β2 is equal to: (A) 37 (B) 58 (C) 68 (D) 92

Tardigrade Admin · Accepted Answer

ax 2-2 bx +15=0 2 α=(2 b / a ), α2=(15/ a ) (α/2)=(15/2 b ) α=(15/ b ) x 2-2 bx +21=0 ((15/ b ))2-2 b ((15/ b ))+21=0 b 2=25 α+β=2 b , α β=21 α2+β2=4 b 2-42 =58

Q. Let $a, b \in R$ be such that the equation $a x^{2} - 2 b x + 15 = 0$ has a repeated root $α$ . If $α$ and $β$ are the roots of the equation $x^{2} - 2 b x + 21 = 0$ , then $α^{2} + β^{2}$ is equal to:

37

58

68

92

$a x^{2} - 2 b x + 15 = 0$
$2 α = \frac{2 b}{a}, α^{2} = \frac{15}{a}$
$\frac{α}{2} = \frac{15}{2 b}$
$α = \frac{15}{b}$
$x^{2} - 2 b x + 21 = 0$
$(\frac{15}{b})^{2} - 2 b (\frac{15}{b}) + 21 = 0$
$b^{2} = 25$
$α + β = 2 b, α β = 21$
$α^{2} + β^{2} = 4 b^{2} - 42$
$= 58$

Q. Let a,b∈R be such that the equation ax2−2bx+15=0 has a repeated root α. If α and β are the roots of the equation x2−2bx+21=0, then α2+β2 is equal to:

37

58

68

92

ax2−2bx+15=0 2α=a2b​,α2=a15​ 2α​=2b15​ α=b15​ x2−2bx+21=0 (b15​)2−2b(b15​)+21=0 b2=25 α+β=2b,αβ=21 α2+β2=4b2−42 =58

Q. Let $a, b \in R$ be such that the equation $a x^{2} - 2 b x + 15 = 0$ has a repeated root $α$ . If $α$ and $β$ are the roots of the equation $x^{2} - 2 b x + 21 = 0$ , then $α^{2} + β^{2}$ is equal to:

$a x^{2} - 2 b x + 15 = 0$
$2 α = \frac{2 b}{a}, α^{2} = \frac{15}{a}$
$\frac{α}{2} = \frac{15}{2 b}$
$α = \frac{15}{b}$
$x^{2} - 2 b x + 21 = 0$
$(\frac{15}{b})^{2} - 2 b (\frac{15}{b}) + 21 = 0$
$b^{2} = 25$
$α + β = 2 b, α β = 21$
$α^{2} + β^{2} = 4 b^{2} - 42$
$= 58$