We can rewrite the given equation as $x^2 + (4m - 2)x + (m + 6) = 0.$ Since there is only one real root, the discriminant must be 0. Hence,
\[(4m - 2)^2 - 4(m + 6) = 0.\]Expanding, we get $16m^2 - 56m + 4 = 0,$ which factors as $4(4m^2 - 14m + 1) = 0.$ By the quadratic formula,
\[m = \frac{14 \pm \sqrt{14^2 - 4 \cdot 4}}{8} = \frac{14 \pm 6 \sqrt{2}}{8} = \frac{7 \pm 3 \sqrt{2}}{4}.\]The positive solution is $\boxed{\frac{7 + 3 \sqrt{2}}{4}}.$
The quadratic equation $x^2+4mx+m = 2x - 6$ has exactly one real root. Find the positive value of $m$.
3 answers
wrong again.
If it has one root, the discriminant is zero.
x^2+4mx+m = 2x - 6
x^2+(4m-2)x+(m-6) = 0
so we need (4m-2)^2-4(m+6) = 0
(2m-1)^2 - (m+6) = 0
4m^2-5m-5 = 0
m = (5±√105)/8
so the positive value is (5+√105)/8
If it has one root, the discriminant is zero.
x^2+4mx+m = 2x - 6
x^2+(4m-2)x+(m-6) = 0
so we need (4m-2)^2-4(m+6) = 0
(2m-1)^2 - (m+6) = 0
4m^2-5m-5 = 0
m = (5±√105)/8
so the positive value is (5+√105)/8
You are right. I apologize for the mistake. The positive value of $m$ is indeed $\boxed{\frac{5+\sqrt{105}}{8}}.$