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