Determine p such that the equation x^{2}+5px+ | vedclass

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

Question

(B) The given quadratic equation is $x^{2}+5px+16=0$. Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1$,$b=5p$,and $c=16$. For the qu

Vedclass · Accepted Answer

$-\frac{8}{5} < p < \frac{8}{5}$

Vedclass · Answer

(B) The given quadratic equation is $x^{2}+5px+16=0$. Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1$,$b=5p$,and $c=16$. For the quadratic equation to have no real roots,the discriminant $D$ must be less than $0$. $D = b^{2}-4ac Substituting the values,we get $(5p)^{2}-4(1)(16) $25p^{2}-64 $25p^{2} $p^{2} Taking the square root on both sides,we get $|p| This implies $-\frac{8}{5} < p < \frac{8}{5}$.

Determine $p$ such that the equation $x^{2}+5px+16=0$ has no real roots.

Similar Questions

Number of equations of the form $ax^2 + bx + 1 = 0$ having real roots,where $a, b \in \{1, 2, 3, 4\}$ is-

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation

If the graph of $y = ax^3 + bx^2 + cx + d$ is symmetric about the line $x = k$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=(x)^{\frac{1}{2}}$
$II.$ $y^{10}-(36)^{5}=0$

If $a + b + c = 0$,$a \neq 0$,$a, b, c \in Q$,then both the roots of the equation $ax^2 + bx + c = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module