Answer:
The values of m for which the equation has one real solution are m = 161/28.
Explanation:
To find all values of m for which the equation (m–4)r^2–7r+7=0 has one real solution, we can use the discriminant of the quadratic equation.
The discriminant is given by the formula: b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In this case, the coefficients are a = (m-4), b = -7, and c = 7.
Calculate the discriminant:
Discriminant = (-7)^2 - 4 * (m-4) * 7
. Simplify the expression:
Discriminant = 49 - 28(m-4)
. Expand the expression inside the parentheses:
Discriminant = 49 - 28m + 112
. Combine like terms:
Discriminant = -28m + 161
To have exactly one real solution, the discriminant must be equal to zero.
. Set the discriminant equal to zero and solve for m:
-28m + 161 = 0
. Move the constant term to the other side:
-28m = -161
. Divide both sides by -28:
m = 161/28