Question
upstudy study bank question image url

.4. For which values of \( p \) will the equation \( x^{2}+x=p \) have no real roots?

Ask by Stephens Welch. in South Africa
Mar 15,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The equation \( x^2 + x = p \) has no real roots when \( p < -\frac{1}{4} \).

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Extra Insights

To determine for which values of \( p \) the equation \( x^{2}+x=p \) has no real roots, we can rewrite the equation as \( x^{2} + x - p = 0 \). The roots of this quadratic equation can be found using the discriminant, which is given by \( D = b^2 - 4ac \). Here, \( a = 1 \), \( b = 1 \), and \( c = -p \). Thus, the discriminant becomes: \[ D = 1^2 - 4(1)(-p) = 1 + 4p. \] For the quadratic equation to have no real roots, the discriminant must be less than zero: \[ 1 + 4p < 0. \] Solving this inequality for \( p \): \[ 4p < -1 \implies p < -\frac{1}{4}. \] Therefore, the equation \( x^{2}+x=p \) will have no real roots for \( p < -\frac{1}{4} \).

Related Questions

Latest Algebra Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy