Examine's 2u horrors (1 Viewer)

[AAEldar]

Can you clarify this step?

Well, since he stated at the start that one of the roots was a then we know that by subbing a into the Polynomial, it will equal zero because it is a root.

 

[Examine]

I'm afraid I'm still not getting it :l

 

[Carrotsticks]

I'm afraid I'm still not getting it :l

Consider the polynomial P(x) = x^2 + 2x + 1.

If I substitute x=-1, then p(-1) = 0 doesn't it? The reason is because x=-1 is a root.

If I have some root and I substitute it into the polynomial from whence it came, then I will get 0.

If a is a root of P(x), then P(a) = 0.

 

[Examine]

So how do you solve it, if there is a k^2 when you are trying to?

 

[SpiralFlex]

I can't be bothered reading the above in its context. Type out whole question.

 

[Examine]

Need help with this question.

The roots of the quadratic equation
x^2-(2k+4)x+(k^2+3k+2)=0
are non-zero and one of the roots is twice the other root, calculate the value of k.

 

[SpiralFlex]

Let the roots be







Now,













But

 

[twinklegal19]

Haha today when I was doing homework, I wrote "Examine 13" instead of "Example 13"...

 

[SpiralFlex]

The influence of BOS....