Let S = x ∈ R : x ≥ 0 & 2 x - 3 + x x - 6 + 6 = 0 . Then S :

2 x - 3 + x x - 6 + 6 = 0 Let x = t > 0 2 t - 3 + t t - 6 + 6 = 0 ⇒ 2 t - 3 = - t 2 - 6 t + 6 ⇒ 2 t - 3 = - t 2 - 6 t + 9 + 3 So the equation 2 solutions.

Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationInequationEasy2 minPYQ_2018

MathematicsEasysingle choice

Let $S = \{x \in R : x \geq 0 & 2 |\sqrt{x} - 3| + \sqrt{x} (\sqrt{x} - 6) + 6 = 0\}$ . Then $S$ :

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Solution:

$2 |\sqrt{x} - 3| + \sqrt{x} (\sqrt{x} - 6) + 6 = 0$

Let $\sqrt{x} = t > 0$

$2 |t - 3| + t (t - 6) + 6 = 0$

$\Rightarrow 2 |t - 3|$ $= - (t^{2} - 6 t + 6)$

$\Rightarrow 2 |t - 3| = - (t^{2} - 6 t + 9) + 3$

So the equation $2$ solutions.

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Inequation

⏱ 2mℹ️ Source: PYQ_2018

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Mathematics - Quadratic Equation Question with Solution | TestHub

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