Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationMaximum & minimum valuesEasy2 minQB

MathematicsEasysingle choice

If $x^{2} + 4 y^{2} - 8 x + 12 = 0$ is satisfied by real values of $x$ and $y$ then ' $y$ ' $\in$

Answer:

Solution:

Rearrange as a quadratic in $x$ : $x^{2} - 8 x + (4 y^{2} + 12) = 0$ .

For real $x$ , the discriminant must be $\geq 0$ .

$y^{2} \leq 1$

This implies $- 1 \leq y \leq 1$ .

Thus, $y \in [- 1, 1]$ .

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Maximum & minimum values

⏱ 2mℹ️ Source: QB

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