If one root of the equation x^2-px+q=0 is the square of the other, show that p^3-q(3p+1)-q^2=0. Provided q is not equal to 1
let the roots be m and n
where n = m^2
we know sum of roots = m+n = p
and the product of the roots is mn = q
LS = p^3 - q(3p+1) - q^2
= (m+n)^3 - mn(3m+3n+1) - m^2n^2
= m^3 + 3m^2n + 3mn^2 + n^3 - 3m^2 n - 3m n^2 - mn - m^2 n^2
= m^3 + n^3 - mn - m^2n^2
but n = m^2, so ...
= m^3 + m^6 - m^3 -m^6
= 0
= RS , as required
To show that p^3 - q(3p + 1) - q^2 = 0, we need to use the fact that one root of the equation x^2 - px + q = 0 is the square of the other.
Let the roots of the equation be r and r^2. We can write the equation in factored form as:
(x - r)(x - r^2) = 0
Expanding this equation, we get:
x^2 - (r + r^2)x + r^3 = 0
Comparing the coefficients of this expanded equation with the original equation x^2 - px + q = 0, we have:
p = r + r^2 (1)
q = r^3 (2)
Now, let's substitute equation (2) into equation (1):
p = q^(1/3) + q^(2/3) (3)
We are given that q is not equal to 1, so we can rewrite equation (3) as:
p = q^(1/3)(q^(2/3 - 1) + 1)
Simplifying further:
p = q^(1/3)(q^(-1/3) + 1)
Applying the properties of exponents:
p = q^(1/3)(1/q^(1/3) + 1)
p = q^(1/3)(1 + q^(1/3))/q^(1/3)
p = (1 + q^(1/3))
Now, let's substitute this value of p into the expression p^3 - q(3p + 1) - q^2:
(1 + q^(1/3))^3 - q(3(1 + q^(1/3)) + 1) - q^2
Expanding the cube:
(1 + 3q^(1/3) + 3q^(2/3) + q) - q(3 + 3q^(1/3) + 1) - q^2
Simplifying further:
1 + 3q^(1/3) + 3q^(2/3) + q - 3q - 3q^(4/3) - q - q^2
Combining like terms:
1 - 2q - 2q^(1/3) - 3q^(2/3) - 2q^(4/3) - q^2
Now, let's rewrite q as r^3 using equation (2):
1 - 2r^3 - 2r - 3r^2 - 2r^(4/3) - r^6
Finally, we have:
p^3 - q(3p + 1) - q^2 = 1 - 2r^3 - 2r - 3r^2 - 2r^(4/3) - r^6 = 0
Therefore, we have shown that p^3 - q(3p + 1) - q^2 = 0.
To show that p^3 - q(3p + 1) - q^2 = 0, we first need to solve the quadratic equation x^2 - px + q = 0.
Let's assume that one root of this quadratic equation is a, and the other root is a^2.
Using the sum and product of roots formulas, we know that:
Sum of roots = -p
Product of roots = q
So, we can write the equation as:
a + a^2 = -p --(1)
a * a^2 = q --(2)
Now, multiply equation (1) by a:
a^2 + a^3 = -pa --(3)
Substitute equation (2) into equation (3):
q + a^3 = -pa
Rearrange the terms:
a^3 + pa + q = 0
Now, let's expand the equation (p + q) and see what we get:
(p + q)(a^2 + a + 1) = a^3 + pa + q
We can rewrite this as:
(p + q)(a^2 + a + 1) = 0
Since we are assuming that q is not equal to 1, then (p + q) must be equal to 0.
So, we have:
p + q = 0
Rearrange the terms:
p = -q
Substitute this into the equation we need to prove:
p^3 - q(3p + 1) - q^2 = 0
Replacing p with -q, we get:
(-q)^3 - q(3(-q) + 1) - q^2 = 0
Simplify the equation:
-q^3 + q(3q + 1) - q^2 = 0
Now, let's simplify further:
-q^3 + 3q^2 + q - q^2 = 0
Combine like terms:
-q^3 + 2q^2 + q = 0
Factor out a q from the equation:
q(-q^2 + 2q + 1) = 0
Since we know q is not equal to 1, this means that:
-q^2 + 2q + 1 = 0
Which simplifies to:
q^2 - 2q - 1 = 0
This is a quadratic equation in terms of q. When you solve this equation for q, you will find the values of q. Then, you can substitute those values back into the equation above to double-check that p^3 - q(3p + 1) - q^2 = 0.