Solving for m with Equal Roots
The problem requires finding the value(s) of m for the quadratic equation $(4 + m)x^2 + (m + 1)x + 1 = 0$ when its roots are equal.
Condition for Equal Roots
A quadratic equation $ax^2 + bx + c = 0$ has equal roots if its discriminant, $D$, is equal to zero. The discriminant is calculated as:
$D = b^2 - 4ac$
For the given equation, we identify the coefficients:
- $a = 4 + m$
- $b = m + 1$
- $c = 1$
Applying the Discriminant Condition
Setting the discriminant to zero:
$D = (m + 1)^2 - 4(4 + m)(1) = 0$
Calculating Values of m
Now, we solve the equation for m:
- Expand the terms: $(m^2 + 2m + 1) - (16 + 4m) = 0$
- Simplify the equation: $m^2 + 2m + 1 - 16 - 4m = 0$
- Combine like terms: $m^2 - 2m - 15 = 0$
- Factor the quadratic equation for m: $(m - 5)(m + 3) = 0$
- Determine the possible values for m: $m - 5 = 0$ or $m + 3 = 0$
- Therefore, the values are: $m = 5$ or $m = -3$.
Conclusion
The values of m for which the equation $(4 + m)x^2 + (m + 1)x + 1 = 0$ has equal roots are $5$ and $-3$.
Similar Questions
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- Solve the following :
$\sqrt{7 + \sqrt{7 + \sqrt{7 + \sqrt{7 + \cdots}}}}$ The sum of two positive numbers is 14 and their product is 45. The positive difference between them is:
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- Which of the quadratic equations below will not have real roots?
Important Questions from Quadratic Equation
For what values of k, the roots of 9x 2 + 8kx + 16 = 0 are real and equal?
- If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of \(\rm \left( 1+\frac{1}{x} \right) \) is equal to :
The nature of the roots of the equation 4x 2 - 2x - 3 = 0.
If the roots of the equation (q – r)x 2+ (r – p)x + (p – q) = 0 are equal, then which of the following is true?
Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is
