If x 2 + 2 a x + 10 − 3 a > 0 for each x ∈ R , then

x 2 + 2 a x + 10 − 3 a > 0∀ x ∈ R ⇒ ( x + a ) 2 − ( a 2 + 10 − 3 a ) > 0∀ x ∈ R ⇒ a 2 + 3 a − 10 < 0 ⇒ ( a + 5 ) ( a − 2 ) < 0 ⇒ − 5 < a < 2

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Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationInequationEasy2 minQB

MathematicsEasysingle choice

If $x^{2} + 2 a x + 10 - 3 a > 0$ for each $x \in R$ , then

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Answer:

Solution:

$x^{2} + 2 a x + 10 - 3 a > 0\forall x \in R$ $\Rightarrow (x + a)^{2} - (a^{2} + 10 - 3 a) > 0\forall x \in R$ $\Rightarrow a^{2} + 3 a - 10 < 0$ $\Rightarrow (a + 5) (a - 2) < 0$ $\Rightarrow - 5 < a < 2$

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Inequation

⏱ 2mℹ️ Source: QB

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