Math, asked by sbajaj544, 11 months ago

Find a quadratic equation whose roots are root 2 and 1/3

Answers

Answered by luckyjoshi615

x^2-(2+1/3)x+2×1/3

=x^2-7/3x+1/3

=3x^2-7x+1

Answered by abdul9838

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$\huge \: Hola \\ \\ Hey \: user \: Here \: is \: Ans. \\ \\ \huge \underline{Question} \\ \\ \underline{Find \: the \: Quadratic \: equation} \\ \\ \huge \underline {Answer} \\ \\ \underline{ \underline{Step \: by \: Step \: explaination} }\\ \\ \\ \\ \\ \\ \huge \bf \: given \: that \\ \\ \bf \: \sqrt{2} \: and \: \frac{1}{3} $

$\bf \: let \: be \\ \\ \bf \alpha = \sqrt{2} \\ \\ \bf \: \beta = \frac{1}{3} \\ \\ \bf \underline{using \: to \: this \: formula} \\ \\ \bf \: (x - \alpha )(x - \beta ) \\ \\ \bf \: (x - \sqrt{2} )(x - \frac{1}{3} ) \\ \\ \bf \: {x}^{2} - \frac{1}{3} x - \sqrt{2} x + \frac{ \sqrt{2} }{3 } \\ \\ \bf \: {x}^{2} - \frac{x}{3 } - \sqrt{2} x + \frac{ \sqrt{2} }{3} \\ \\ \bf \: {x}^{2} - \frac{ x - 3 \sqrt{2}x }{3} + \frac{ \sqrt{2} }{3} \\ \\ \bf \: {x}^{2} + \frac{ \sqrt{2} - 3 \sqrt{2}x - x }{3} \\ \\ \bf \: \frac{ {3x}^{2} - 3 \sqrt{2} x - x + \sqrt{2} }{3} \\ \\ \bf \: \frac{1}{3} (3 {x}^{2} - 3 \sqrt{2} x - x + \sqrt{2} ) \: \: ans$

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