Question
upstudy study bank question image url

The reots of the equation \( f(x)=0 \) is \( x=\frac{4 \pm \sqrt{6-4 m /-m+5)}}{2 m} \) Determine the valaes of \( m \) for which the roots are non-real

Ask by Davey Reid. in South Africa
Feb 24,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The roots are non-real when \( \frac{3}{2} < m < 5 \).

Solution

We are given that the roots of f(x) = 0 are   x = [4 ± √( (6 – 4m)/(–m + 5) )] / (2m). For the roots to be non‐real the expression under the square root (the discriminant) must be negative. In other words, we require   (6 – 4m)/(–m + 5) < 0. Step 1. Rewrite the inequality in a slightly simpler form. Notice that   –m + 5 = 5 – m. So the inequality becomes   (6 – 4m)/(5 – m) < 0. Step 2. Find the critical values by setting the numerator and denominator equal to zero. • Numerator: 6 – 4m = 0 ⇒ 4m = 6 ⇒ m = 3/2. • Denominator: 5 – m = 0 ⇒ m = 5. Also note that m = 5 must be excluded since it would make the denominator zero (and would also make the expression under the square root undefined). Additionally, since m appears as a coefficient in the denominator of the entire root expression (x = …/(2m)), one might exclude m = 0 to maintain a quadratic formula—but our focus here is determining when the discriminant is negative. Step 3. Determine the sign of the expression (6 – 4m)/(5 – m) in the intervals determined by m = 3/2 and m = 5. The intervals to test are:  (a) m < 3/2,  (b) 3/2 < m < 5,  (c) m > 5. Test a value in each region: (a) For m = 0 (which is less than 3/2):   Numerator: 6 – 4(0) = 6  (positive).   Denominator: 5 – 0 = 5  (positive).   So (6 – 4m)/(5 – m) > 0. (b) For m = 2 (which lies between 3/2 and 5):   Numerator: 6 – 4(2) = 6 – 8 = –2  (negative).   Denominator: 5 – 2 = 3  (positive).   So (6 – 4m)/(5 – m) < 0. (c) For m = 6 (which is greater than 5):   Numerator: 6 – 4(6) = 6 – 24 = –18  (negative).   Denominator: 5 – 6 = –1  (negative).   So (6 – 4m)/(5 – m) > 0 (since a negative divided by a negative is positive). Step 4. Conclusion The expression (6 – 4m)/(5 – m) is negative only on the interval   3/2 < m < 5. Thus, the roots of the equation are non‐real if and only if m is in the open interval (3/2, 5). Answer: The roots are non‐real for all m satisfying 3/2 < m < 5.

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Beyond the Answer

To determine the values of \( m \) for which the roots of the equation \( f(x)=0 \) are non-real, we need to analyze the expression under the square root (the discriminant). The roots are non-real when the discriminant is less than zero. Given \( x=\frac{4 \pm \sqrt{6-4 m / (-m+5)}}{2 m} \), the discriminant is \( 6-4m / (-m+5) \). For the roots to be non-real, we must solve: \[ 6 - \frac{4m}{-m + 5} < 0 \] First, rearranging gives: \[ -\frac{4m}{-m + 5} < -6 \] Multiplying through by \(-(-m + 5)\) (reversing the inequality) we have: \[ 4m > 6(-m + 5) \] Distributing: \[ 4m > -6m + 30 \] Combining like terms yields: \[ 10m > 30 \implies m > 3 \] Thus, the values of \( m \) for which the roots are non-real are \( m > 3 \).

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