Mathematics
Find the value of m for which the equation (m + 4)x2 + (m + 1)x + 1 = 0 has real and equal roots.
Quadratic Equations
2 Likes
Answer
Since, equation has equal roots, D = 0.
∴ b2 - 4ac = 0
⇒ (m + 1)2 - 4(m + 4)(1) = 0
⇒ m2 + 1 + 2m - 4m - 16 = 0
⇒ m2 - 2m - 15 = 0
⇒ m2 - 5m + 3m - 15 = 0
⇒ m(m - 5) + 3(m - 5) = 0
⇒ (m + 3)(m - 5) = 0
⇒ (m + 3) = 0 or (m - 5) = 0
⇒ m = -3 or m = 5.
Hence, m = -3 or 5.
Answered By
1 Like
Related Questions
Solve :
(a + b)2x2 - (a + b)x - 6 = 0; a + b ≠ 0.
Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
x2 + 2(m - 1)x + (m + 5) = 0
Find the value of k for which equation 4x2 + 8x - k = 0 has real roots.
If -2 is a root of the equation 3x2 + 7x + p = 1, find the value of p. Now find the value of k so that the roots of the equation x2 + k(4x + k - 1) + p = 0 are equal.