Find the values of p if the equation $x^{2} - x + p = 0$ does not possess real roots.
(A) p > $\frac{1}{8}$
(B) p > $\frac{1}{4}$
(C) p > $\frac{1}{2}$
(D) p > 1
$[1]$

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Solution

Option B is the correct answer.

On comparing the given equation $x^{2} - x + p = 0$ with standard form $a x^{2} + b x + c = 0$ , we get

a = 1, b = -1 and c = p

We know that when a quadratic equation has no real roots, discriminant, D < 0.

$\Rightarrow b^{2} - 4 a c < 0$

$\Rightarrow (- 1)^{2} - 4 (1) (p) < 0$

$\Rightarrow 1 - 4 p < 0$

$\Rightarrow 4 p > 1$

$\Rightarrow p > \frac{1}{4}$

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Find the values of p if the equation x2−x+p=0 does not possess real roots. (A) p > 18 (B) p > 14 (C) p > 12 (D) p > 1 [1]

Option B is the correct answer. On comparing the given equation x2−x+p=0 with standard form ax2+bx+c=0, we get a = 1, b = -1 and c = p We know that when a quadratic equation has no real roots, discriminant, D < 0. ⇒b2−4ac<0 ⇒(−1)2−4(1)(p)<0 ⇒1−4p<0 ⇒4p>1 ⇒p>14

Find the values of p if the equation $x^{2} - x + p = 0$ does not possess real roots.
(A) p > $\frac{1}{8}$
(B) p > $\frac{1}{4}$
(C) p > $\frac{1}{2}$
(D) p > 1
$[1]$