Find the value of π if one root of the quadratic equation π₯2 β ππ₯ + 8 = 0 is the square of the other.
Let the roots of the quadratic equation be π and π, where π is the square of π. Then we have:
π + π = π (sum of roots is π)
ππ = 8 (product of roots is 8)
Substituting π = π2 into the first equation, we get:
π + π2 = π
Simplifying this equation, we get:
π2 + π β π = 0
This is a quadratic equation in π with a = 1, b = 1 and c = -π. We know that the roots of this equation are π and π = π2. Therefore, by Vieta's formulas:
π + π = -b/a
Substituting the values of π and a, we get:
π + π2 = -1/1
π2 + π + 1 = 0
The discriminant of this equation is:
π2 β 4ππ = 1 β 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 and π, where π is the square of π.
So, π = π^2
According to the quadratic equation, the sum of the roots (π + π) is equal to the negation of the linear coefficient of π₯ divided by the leading coefficient.
π + π = -(-π) / 1 = π
We also know that the product of the roots (π * π) is equal to the constant term divided by the leading coefficient.
π * π = 8 / 1 = 8
Substituting π = π^2 in the equation π * π = 8, we get:
π * π^2 = 8
π^3 = 8
Taking the cube root of both sides:
π = β8
We know that 8 can be expressed as 2^3, so:
π = β(2^3)
π = 2^(3/3)
π = 2^1
π = 2
Now we can find the value of π:
π = π + π
= 2 + π^2
= 2 + (2)^2
= 2 + 4
= 6
Therefore, the value of π is 6.