Question
If -5 is a root of the quadratic equation $2 x^2+p x-15=0$, then:(a) $\mathrm{p}=3$(b) $\mathrm{p}=5$(c) $p=7$(d) $\mathrm{p}=1$
Step 1
Step 1: Since -5 is a root of the quadratic equation \(2x^2 + px - 15 = 0\), we can substitute \(x = -5\) into the equation. Show more…
Show all steps
Your feedback will help us improve your experience
Anas Venkitta and 57 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If the equation $(3 x)^{2}+\left(27 \times 3^{1 / p}-15\right) x+4=0$ has equal roots then $p$ is (a) 0 (b) 2 (c) $-\frac{1}{2}$ (d) 1
If one of the roots of the equation $x^{2}-(p+1) x+p^{2}+$ $p-8=0$ is greater than 2 and the other root is smaller than 2 , then $p$ is such that (A) $-\frac{11}{3}<p<3$ (B) $-2<p<3$ (C) $2<p<3$ (D) None of these
If the equation $x^{2}-15-m(2 x-8)=0$ has equal roots then the value of $m$ can be (a) 15 or 8 (b) 0 or 2 (c) 4 or 8 (d) 5 or 3
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD