how do prove that this equation has rational roots for all rational values of K 3x^2+kx-k=3x
first write your equation in standard quadratic form
3x^2 + kx - 3x - k = 0
3x^2 + (k-3)x - k = 0
so
a=3
b=k-3
c=-k
the discriminant is b^2 - 4ac
= (k-3)^2 - 4(3)(-k)
= k^2 - 6k + 9 + 12k
= k^2 + 6k + 9
= (k+3)^2
for rational roots, (k+3)^2 ≥ 0
but the square of anything is always ≥ 0
so your equation has rational roots for any value of k
(no matter what you put in for k, (k+3)^2 is > 0 )
To prove that the equation 3x^2 + kx - k = 3x has rational roots for all rational values of K, we can follow these steps:
Step 1: Assume that K is a rational number.
Step 2: Rewrite the equation in quadratic form by moving all terms to one side: 3x^2 + (k - 3)x - k = 0.
Step 3: Use the quadratic formula to find the roots of the quadratic equation. The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the roots can be found using the formula: x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 3, b = (k - 3), and c = -k. Substituting these values into the quadratic formula, we get x = (-(k - 3) ± √((k - 3)^2 - 4(3)(-k))) / (2(3)).
Step 4: Simplify the quadratic formula expression:
x = (-(k - 3) ± √(k^2 - 6k + 9 + 12k)) / 6
x = (-(k - 3) ± √(k^2 + 6k + 9)) / 6
x = (3 - k ± √((k + 3)^2)) / 6
x = (3 - k ± (k + 3)) / 6
x = 6/6, -k/6
Step 5: Simplify the roots:
x = 1, -k/6
Since K is assumed to be a rational number, both roots of the equation are rational. Hence, the equation 3x^2 + kx - k = 3x has rational roots for all rational values of K.
To prove that the equation 3x^2 + kx - k = 3x has rational roots for all rational values of K, we need to show that the equation can be factored into linear factors with rational coefficients.
First, let's rewrite the equation in a standard quadratic form by moving all the terms to one side:
3x^2 + kx - k - 3x = 0
Combining like terms, we get:
3x^2 + (k - 3)x - k = 0
Now, to prove that this equation has rational roots for all rational values of K, we need to show that its discriminant is a perfect square.
The discriminant of a quadratic equation in the form ax^2 + bx + c = 0 is given by Δ = b^2 - 4ac. In our equation, a = 3, b = (k - 3), and c = -k.
Substituting the values into the discriminant formula, we have:
Δ = (k - 3)^2 - 4(3)(-k)
Expanding the expression, we get:
Δ = k^2 - 6k + 9 + 12k
Simplifying, we have:
Δ = k^2 + 6k + 9
Now, we need to determine whether Δ is a perfect square for all rational values of K. One way to check this is by factoring Δ and making sure it factors into the product of two equal rational numbers.
Factoring Δ, we get:
Δ = (k + 3)(k + 3)
Simplifying, we have:
Δ = (k + 3)^2
Since (k + 3)^2 is always a perfect square for any value of K, we have proved that the discriminant of our quadratic equation is always a perfect square. Therefore, the quadratic equation 3x^2 + kx - k = 3x has rational roots for all rational values of K.