Quadratics: Quadratic Equations

[James400]

Consider the quadratic equation x^2+px+2p=0

a. Find the discriminant.

b. Find the values of p for which there are 2 solutions.

c. Find the values of p for which there are no solutions.

d. Find the value of p for which there is 1 solution.

Please show working out! Thanks.

 

[Joppy]

Please show working out! Thanks.

How about your working out! :). How can we help if we don't know what you are struggling with :p.

 

[James400]

For a. Find the discriminant:
=b^2-4ac
=p^2-8p

That's is as far as I have gotten, although I think that is right for a. correct me if I am wrong. So I just need to solve the others, although I have no idea how to do so...

 

[James400]

Consider the quadratic equation x^2+px+2p=0

a. Find the discriminant.

b. Find the values of p for which there are 2 solutions.

c. Find the values of p for which there are no solutions.

d. Find the value of p for which there is 1 solution.

Please show working out! Thanks.

Is anyone able to confirm whether a. is p^2-4-8p and help solve b through to d?

 

[Greg]

Ok, let's start from scratch.

$$ax^2+bx+c=0$$

$$a(x^2+\frac bax)=-c$$

$$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}=-c$$

$$a\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a}-c$$

$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac ca$$

$$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$$

$$x=-\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$$

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Now plug your values for $a$, $b$ and $c$ into $b^2-4ac$.

Where is $b^2-4ac>0$? Less than $0$? Equal to $0$? Do you understand the relevance of these relationships?