Find the value of 𝑘 if one root of the quadratic equation 𝑥2 − 𝑘𝑥 + 8 = 0 is the square of the other.

Answer 1

Let the roots of the quadratic equation be 𝑎 and 𝑏, where 𝑏 is the square of 𝑎. Then we have:

𝑎 + 𝑏 = 𝑘 (sum of roots is 𝑘)
𝑎𝑏 = 8 (product of roots is 8)

Substituting 𝑏 = 𝑎2 into the first equation, we get:

𝑎 + 𝑎2 = 𝑘

Simplifying this equation, we get:

𝑎2 + 𝑎 − 𝑘 = 0

This is a quadratic equation in 𝑎 with a = 1, b = 1 and c = -𝑘. We know that the roots of this equation are 𝑎 and 𝑏 = 𝑎2. Therefore, by Vieta's formulas:

𝑎 + 𝑏 = -b/a

Substituting the values of 𝑏 and a, we get:

𝑎 + 𝑎2 = -1/1

𝑎2 + 𝑎 + 1 = 0

The discriminant of this equation is:

𝑏2 − 4𝑎𝑐 = 1 − 4(1)(1) = -3

Since the discriminant is negative, the quadratic equation has no real roots. Therefore, there is no real value of 𝑘 that satisfies the condition that one root is the square of the other.