Question

$\huge\blue{\bold{\underline{\underline{Question:-}}}}$

If α , β are the roots of the quadratic equation 3x² + 2x - 1 = 0 , Then find the quadratic equation whose roots are

(i) α + k , β + k

(ii) α - k , β - k

[ Question related to Transfirmed Quadratic equations . NO SPAM ] ​

Answer 1

GIVEN :–

• A quadratic equation 3x² + 2x - 1 = 0 which have two roots α and β.

TO FIND :–

• Quadratic equation whose roots are –

(i) α + k , β + k

(ii) α - k , β - k

SOLUTION :–

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$\bf \implies 3 {x}^{2} + 2x - 1 = 0$

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• Splitting Middle term –

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$\bf \implies 3 {x}^{2} + 3x - x - 1 = 0$

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$\bf \implies 3x(x + 1) -(x + 1) = 0$

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$\bf \implies (3x - 1)(x + 1) = 0$

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$\bf \implies x = \dfrac{1}{3}, - 1$

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• Hence –

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$\bf \implies \large { \boxed{ \bf \alpha = \dfrac{1}{3}}} \: \: and \: \:{ \boxed{ \bf \beta = - 1}}$

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(i) Quadratic equation which roots are α + k , β + k.

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• We know that a quadratic equation –

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$\bf \: \: \: \bigstar \: \: \: \large{ \boxed{ \bf {x}^{2} - (sum \: \: of \: \: roots)x + (product \: \: of \: \: roots) = 0}}$

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▪︎Now –

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• Sum of roots = (α + k) + (β + k) = (⅓ + k) + (-1 + k) = 2k - ⅔

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• Product of roots = (α + k).(β + k) = (⅓ + k).(-1 + k) = k² - ⅔ k - ⅓

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• So that , Quadratic equation –

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$\bf \implies{ \boxed{ \bf {x}^{2} - \left(2k - \dfrac{2}{3} \right)x + \left( {k}^{2} - \dfrac{2}{3}k - \dfrac{1}{3} \right) = 0}}$

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(ii) Quadratic equation which roots are α - k , β - k.

ㅤ

• We know that a quadratic equation –

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$\bf \: \: \: \bigstar \: \: \: \large{ \boxed{ \bf {x}^{2} - (sum \: \: of \: \: roots)x + (product \: \: of \: \: roots) = 0}}$

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▪︎Now –

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• Sum of roots = (α - k) + (β - k) = (⅓ - k) + (-1 - k) = - ⅔

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• Product of roots = (α - k).(β - k) = (⅓ - k).(-1 - k) = k² + ⅔ k - ⅓

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• So that , Quadratic equation –

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$\bf \implies{ \boxed{ \bf {x}^{2} + \dfrac{2}{3} x + \left( {k}^{2} + \dfrac{2}{3}k - \dfrac{1}{3} \right) = 0}}$