The zeros of the polynomial x2-1-6 x-2 are a -3- 4 b -3-2- 4-3 c -4-3- 3-2 d none of these

Polynomial : x^2+1/6 x 2 Comparing with the standard equation, we get a = 1, b = 1/6 , c = 2 To find zero, equate it with 0 = gt; x^2+1/6 x 2=0 Multip ...

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Solving a Quadratic Equation by Factorization Method

The zeros of ...

Question

The zeros of the polynomial $x^{2} + \frac{1}{6} x - 2$ are

(a) -3, 4 (b) $\frac{- 3}{2}, \frac{4}{3}$ (c) $\frac{- 4}{3}, \frac{3}{2}$ (d) none of these

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Solution

Polynomial : $x^{2} + \frac{1}{6} x - 2$

Comparing with the standard equation, we get

a = 1, b = $\frac{1}{6}$ , c = -2

To find zero, equate it with 0

=> $x^{2} + \frac{1}{6} x - 2$ =0

Multiply by 6
=> $6 x^{2} + x - 12 = 0$
=> $6 x^{2} + 9 x - 8 x - 12 = 0$
=> $3 x (2 x + 3) - 4 (2 x + 3) = 0$
=> $(2 x + 3) (3 x - 4) = 0$
$2 x + 3 = 0$ Or $3 x - 4 = 0$
=> x = $\frac{- 3}{2} o r \frac{4}{3}$
so option b is correct

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Q. If α, β are the zeros of the polynomial x² + 6x + 2, then

(\frac{1}{α} + \frac{1}{β}) = ?

(a) 3
(b) −3
(c) 12
(d) −12

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The zeros of the polynomial x2+16x−2 are (a) -3, 4 (b) −32,43 (c) −43,32 (d) none of these

The zeros of the polynomial $x^{2} + \frac{1}{6} x - 2$ are

(a) -3, 4 (b) $\frac{- 3}{2}, \frac{4}{3}$ (c) $\frac{- 4}{3}, \frac{3}{2}$ (d) none of these