If both roots of the equation x^2 - (p - 4)x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the quadratic equation $x^2 - (p - 4)x + 2e^{2 \ln p} - 4 = 0$. Since $e^{2 \ln p} = e^{\ln p^2} = p^2$,the equation becomes $x^2 - (p - 4)x

Vedclass · Accepted Answer

$\left( \sqrt{2}, 4 \right)$

Vedclass · Answer

(B) Given the quadratic equation $x^2 - (p - 4)x + 2e^{2 \ln p} - 4 = 0$. Since $e^{2 \ln p} = e^{\ln p^2} = p^2$,the equation becomes $x^2 - (p - 4)x + 2p^2 - 4 = 0$. For both roots to be negative,the following conditions must be satisfied: $1.$ Discriminant $D \ge 0$: $(p - 4)^2 - 4(2p^2 - 4) \ge 0 \implies p^2 - 8p + 16 - 8p^2 + 16 \ge 0 \implies -7p^2 - 8p + 32 \ge 0 \implies 7p^2 + 8p - 32 \le 0$. Solving $7p^2 + 8p - 32 = 0$,we get $p = \frac{-8 \pm \sqrt{64 - 4(7)(-32)}}{14} = \frac{-8 \pm \sqrt{64 + 896}}{14} = \frac{-8 \pm \sqrt{960}}{14} = \frac{-8 \pm 8\sqrt{15}}{14} = \frac{-4 \pm 4\sqrt{15}}{7}$. $2.$ Sum of roots $$3.$ Product of roots $> 0$: $2p^2 - 4 > 0 \implies p^2 > 2 \implies p > \sqrt{2}$ or $p Also,for $\ln p$ to be defined,$p > 0$. Combining $p > \sqrt{2}$ and $p < 4$,we get $p \in (\sqrt{2}, 4)$.

If both roots of the equation $x^2 - (p - 4)x + 2e^{2 \ln p} - 4 = 0$ are negative,then in which interval does $p$ lie?

Similar Questions

If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation:

The value of $a$ for which the quadratic equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ possesses roots with opposite signs,lies in

Let $N$ be the number of quadratic equations of the form $ax^2 + bx + c = 0$ with coefficients $a, b, c \in \{0, 1, 2, \dots, 9\}$ such that $0$ is a solution of each equation. Then the value of $N$ is

If the equation $x^2-3ax+a^2-2a-K=0$ has different real roots for every rational number $a$,then $K$ lies in the interval

The number of distinct real solutions for the equation $|x^2+2x-8|+x-2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module