Find the value of π‘˜ if one root of the quadratic equation π‘₯2 βˆ’ π‘˜π‘₯ + 8 = 0 is the square of the other.

Let the roots of the quadratic equation be π‘Ž and 𝑏, where 𝑏 is the square of π‘Ž. Then we have:

π‘Ž + 𝑏 = π‘˜ (sum of roots is π‘˜)
π‘Žπ‘ = 8 (product of roots is 8)

Substituting 𝑏 = π‘Ž2 into the first equation, we get:

π‘Ž + π‘Ž2 = π‘˜

Simplifying this equation, we get:

π‘Ž2 + π‘Ž βˆ’ π‘˜ = 0

This is a quadratic equation in π‘Ž with a = 1, b = 1 and c = -π‘˜. We know that the roots of this equation are π‘Ž and 𝑏 = π‘Ž2. Therefore, by Vieta's formulas:

π‘Ž + 𝑏 = -b/a

Substituting the values of 𝑏 and a, we get:

π‘Ž + π‘Ž2 = -1/1

π‘Ž2 + π‘Ž + 1 = 0

The discriminant of this equation is:

𝑏2 βˆ’ 4π‘Žπ‘ = 1 βˆ’ 4(1)(1) = -3

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

Let's assume the roots of the quadratic equation are π‘Ž and 𝑏, where 𝑏 is the square of π‘Ž.

So, 𝑏 = π‘Ž^2

According to the quadratic equation, the sum of the roots (π‘Ž + 𝑏) is equal to the negation of the linear coefficient of π‘₯ divided by the leading coefficient.

π‘Ž + 𝑏 = -(-π‘˜) / 1 = π‘˜

We also know that the product of the roots (π‘Ž * 𝑏) is equal to the constant term divided by the leading coefficient.

π‘Ž * 𝑏 = 8 / 1 = 8

Substituting 𝑏 = π‘Ž^2 in the equation π‘Ž * 𝑏 = 8, we get:

π‘Ž * π‘Ž^2 = 8

π‘Ž^3 = 8

Taking the cube root of both sides:

π‘Ž = βˆ›8

We know that 8 can be expressed as 2^3, so:

π‘Ž = βˆ›(2^3)

π‘Ž = 2^(3/3)

π‘Ž = 2^1

π‘Ž = 2

Now we can find the value of π‘˜:

π‘˜ = π‘Ž + 𝑏
= 2 + π‘Ž^2
= 2 + (2)^2
= 2 + 4
= 6

Therefore, the value of π‘˜ is 6.