Question

If α,β are the zeros of the quadratic polynomial 6x2+x-2,then find 1/α+1/β​

Answer 1

$6 {x}^{2} + x - 2 \\ = > 6 {x}^{2} + 4x - 3x - 2 \\ = > 2x(3x + 2) - 1(3x + 2) \\ = > (2x - 1)(3x + 2) \\ = > x = \frac{1}{2} \: and \: \frac{ - 2}{3} \\ \alpha = \frac{1}{2} \: and \: \beta = \frac{ - 2}{3} \\ according \: to \: question \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{1}{ \frac{1}{2} } + \frac{1}{ \frac{ - 2}{3} } \\ = > 2 + ( \frac{ - 3}{2} ) \\ = > 2 - \frac{3}{2} \\ = > \frac{4 - 3}{2} \\ = > \frac{1}{2}$

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Answer 2

Given: α and β are the zeros of the quadratic polynomial 6x² + x - 2.

To find: The value of 1/α + 1/β.

Answer:

Let's suppose ax² + bx + c were a quadratic polynomial.

$\sf The\ sum\ of\ its\ zeros\ would\ be\ given\ by\ \dfrac{-b}{a}.\\\\\\The\ product\ of\ its\ zeros\ would\ be\ given\ by\ \dfrac{c}{a}.$

From the given equation,

• a = 6
• b = 1
• c = -2

Since α and β are the zeros,

$\implies\ \sf \alpha\ +\ \beta\ =\ \dfrac{-1}{6}\\\\\\\implies\ \alpha\ \times\ \beta\ =\ \dfrac{-2}{6}\ =\ \dfrac{-1}{3}$

$\sf \dfrac{1}{\alpha}\ +\ \dfrac{1}{\beta}\ =\ \dfrac{\alpha\ +\ \beta}{\alpha\beta}$

Using the values we have,

$\dfrac{\alpha\ +\ \beta}{\alpha\beta}\ =\ \dfrac{\frac{-1}{6}}{\frac{-1}{3}}\ =\ \dfrac{-1}{6}\ \times\ \dfrac{-3}{1}\ =\ \dfrac{1}{2}\\\\\\\therefore\ \bf \dfrac{1}{\alpha}\ +\ \dfrac{1}{\beta}\ =\ \dfrac{1}{2}.$