For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution? express your answer as a common fraction.

(x+4)(x+1) = m+2x
x² + 5x + 4 − m − 2x = 0
x² + 3x + (4−m) = 0

Quadratic equation has exactly one real solution
when discriminant = 0

b² − 4ac = 0
3² − 4(1)(4−m) = 0
9 − 16 + 4m = 0
−7 + 4m = 0
4m = 7
m = 7/4



Second problem:

Line y = 3 intersects graph of y = 4x² + x − 1 at point A and B

y-coordinates of points A and B = 3, since they are both on line y = 3. We need to find x-coordinates

4x² + x − 1 = 3
4x² + x − 4 = 0

x = (−1 ± √(1−4(4)(−4))) / 8
x = (−1±√65) / 8

AB = (−1+√65)/8 − (−1−√65)/8 = 2√65/8 = √65/4 = √(65/16)

m−n = 49

I have take the answer from internet, I hope it help

step-by-step explanation:

no solution.