Question 10
One of the zeroes of the polynomial is $2 x^{2} + 7 x - 4$ is
A) $2$
B) $\frac{1}{2}$
C) $- \frac{1}{2}$
D) $- 2$

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Solution

The answer is B.
Let $p (x) = 2 x^{2} + 7 x - 4$
$= 2 x^{2} + 8 x - x - 4$ [By splitting middle term]
$= 2 x (x + 4) - 1 (x + 4)$
$= (2 x - 1) (x + 4)$
For zeroes of p(x), put p(x)=0.
$∴ (2 x - 1) (x + 4) = 0$
$\Rightarrow 2 x - 1 = 0$ and $x + 4 = 0$
$\Rightarrow x = \frac{1}{2}$ and $x = - 4$
Hence, one of the zeroes of the polynomial p(x) is $\frac{1}{2}$ .

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Q. Question 10
One of the zeroes of the polynomial is

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A) $2$
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C) $- \frac{1}{2}$
D) $- 2$

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2 x^{2} + 7 x - 4

If α, β are zeroes of the polynomial f(x) = 2x² + 5x - 10, then value of αβ is

The zeros of the polynomial $p (x) = 2 x^{2} + 7 x - 4$ are
(a) $4, \frac{- 1}{2}$
(b) $4, \frac{1}{2}$
(c) $- 4, \frac{1}{2}$
(d) $- 4, \frac{- 1}{2}$

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Roots of Quadratic Equation

Standard X Mathematics

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Question 10 One of the zeroes of the polynomial is 2x2+7x–4 is A) 2 B) 12 C) −12 D) −2

The answer is B. Let p(x)=2x2+7x−4 =2x2+8x–x–4 [By splitting middle term] =2x(x+4)−1(x+4) =(2x–1)(x+4) For zeroes of p(x), put p(x)=0. ∴ (2x−1)(x+4)=0 ⇒ 2x−1=0 and x+4=0 ⇒ x=12 and x=−4 Hence, one of the zeroes of the polynomial p(x) is 12.

Question 10
One of the zeroes of the polynomial is $2 x^{2} + 7 x - 4$ is
A) $2$
B) $\frac{1}{2}$
C) $- \frac{1}{2}$
D) $- 2$