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# Help with Discriminant(C1)

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1. I have this C1 discriminant question that I can't do, anyone know how to do it?
Prove that the roots of the equation kx^2+(2k+4)x+8 are real for all values of K.
x^2=x squared
I have this C1 discriminant question that I can't do, anyone know how to do it?
Prove that the roots of the equation kx^2+(2k+4)x+8 are real for all values of K.
x^2=x squared
So show that for all

3. Boring part

For the equation ax^2 + bx + c = 0, the discriminant is b^2-4ac, and if this number is positive then there are two real roots. So you can figure out what a, b and c are in this particular equation (don't let the ks put you off, the process is the same), then show that this must be positive for real k.

Interesting part

Why is this true?

Because when you complete the square, you get

x^2 + bx/a + c/a = 0

(x + b/2a)^2 - b^2/4a^2 + c/a = 0

(x + b/2a)^2 = b^2/4a^2 - c/a

= b^2/4a^2 - 4ac/4a^2
= (b^2 - 4ac)/(4a^2)

So

(x + b/2a)^2 = (b^2 - 4ac)/(4a^2)
4a^2*(x + b/2a)^2 = (b^2 - 4ac)
[2a*(x + b/2a)]^2 = b^2 - 4ac

2a*(x + b/2a) = +/- sqrt(b^2 - 4ac)

That last line tells you why it is true. If b^2-4ac is positive then its square root, the RHS of this equation, is real and we have real solutions for x. If it is negative then the RHS is an imaginary number, and we have no real solutions. So b^2 - 4ac is the key.
4. Thank you
(Original post by mik1a)
Boring part

For the equation ax^2 + bx + c = 0, the discriminant is b^2-4ac, and if this number is positive then there are two real roots. So you can figure out what a, b and c are in this particular equation (don't let the ks put you off, the process is the same), then show that this must be positive for real k.

Interesting part

Why is this true?

Because when you complete the square, you get

x^2 + bx/a + c/a = 0

(x + b/2a)^2 - b^2/4a^2 + c/a = 0

(x + b/2a)^2 = b^2/4a^2 - c/a

= b^2/4a^2 - 4ac/4a^2
= (b^2 - 4ac)/(4a^2)

So

(x + b/2a)^2 = (b^2 - 4ac)/(4a^2)
4a^2*(x + b/2a)^2 = (b^2 - 4ac)
[2a*(x + b/2a)]^2 = b^2 - 4ac

2a*(x + b/2a) = +/- sqrt(b^2 - 4ac)

That last line tells you why it is true. If b^2-4ac is positive then its square root, the RHS of this equation, is real and we have real solutions for x. If it is negative then the RHS is an imaginary number, and we have no real solutions. So b^2 - 4ac is the key.

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