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Relation between Roots and Coefficients for Quadratic

Let x 2- ax +...

Question

Let $x^{2} - a x + b = 0$ , where $a, b \in R$ be a quadratic equation such that the roots are opposite in sign and the magnitude of one root is twice the other. Then which of the following options is/are always true ?

2 b^{2} + a = 0

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2 a^{2} + b = 0

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b < 0

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b > 0

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Solution

The correct options are
B $2 a^{2} + b = 0$
C $b < 0$
$x^{2} - a x + b = 0 . . . (1)$
Roots are opposite in sign. This means the roots are real.
Let $α, β$ be the roots.
$α + β = a, α β = b; a, b \neq 0$
Let $α = - 2 β$
$\Rightarrow β = - a$
$∴ α = 2 a$

$\Rightarrow α β = b = - 2 a^{2} < 0$

Now, substituting either $α = 2 a$ or $β = - a$ in equation $(1),$ we get
$2 a^{2} + b = 0$

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Relation between Roots and Coefficients for Quadratic

Standard XII Mathematics

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Let x2−ax+b=0, where a,b∈R be a quadratic equation such that the roots are opposite in sign and the magnitude of one root is twice the other. Then which of the following options is/are always true ?

The correct options are B 2a2+b=0 C b<0x2−ax+b=0 ...(1) Roots are opposite in sign. This means the roots are real. Let α,β be the roots. α+β=a, αβ=b; a,b≠0 Let α=−2β ⇒β=−a ∴α=2a ⇒αβ=b=−2a2<0 Now, substituting either α=2a or β=−a in equation (1), we get 2a2+b=0

Let $x^{2} - a x + b = 0$ , where $a, b \in R$ be a quadratic equation such that the roots are opposite in sign and the magnitude of one root is twice the other. Then which of the following options is/are always true ?