3x^2 kx-k=3x prove that this equation has rational roots for all ration values of K thanks
You are missing a + or - sign in there.
Its 3x^2 kx-k=3x prove that this equation has rational roots for all rational values of K.
Sorry.
3x^2 kx-k
makes no sense. Is it
3x^2+kx-k
or
3x^2-kx-k
?
As it stands, since a b means a*b, you have
3x^2*kx-k = 3k
3kx^3-k = 3x
3kx^3-3x-k = 0
which clearly does not work.
3x^2 kx-k=3x
It's 3x^2+kx-k=3k
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 discriminant of the equation is a perfect square for any rational value of k.
First, let's rewrite the equation in the standard form Ax^2 + Bx + C = 0:
3x^2 + (k-3)x - k = 0
The discriminant of a quadratic equation is given by the formula D = B^2 - 4AC. In this case, A = 3, B = (k-3), and C = -k. Substituting these values, we have:
D = (k-3)^2 - 4(3)(-k)
= (k^2 - 6k + 9) + (12k)
= k^2 + 6k + 9 + 12k
= k^2 + 18k + 9
To prove that the equation has rational roots for all rational values of k, we need to show that the discriminant D is a perfect square for any rational value of k. In other words, we need to show that D is a perfect square when k is any rational number.
For D to be a perfect square, its square root must be an integer. Let's take the square root of D:
√D = √(k^2 + 18k + 9)
To simplify this expression, we need to find the factors of k^2 + 18k + 9. We can do this by factoring the quadratic expression:
k^2 + 18k + 9 = (k + 9)(k + 1)
Now, let's substitute back into our expression for √D:
√D = √((k + 9)(k + 1))
At this point, we can see that √D will be a rational number if and only if (k + 9)(k + 1) is a perfect square. Since k is rational, both k + 9 and k + 1 are rational.
Hence, for any rational value of k, the expression (k + 9)(k + 1) can be written as the square of a rational number, which means the equation has rational roots for all rational values of k.
Therefore, we have proven that the equation 3x^2 kx-k = 3x has rational roots for all rational values of k.