Quadratic functions question. Need help! (1 Viewer)

clue444 · Jul 9, 2009

Find the relationship between a and b in the equation ax<sup>2</sup> + bx + c when the:

1) larger root exceeds the smaller root by 1

2) roots are in the ratio 2:3

can someone help me?

kurt.physics · Jul 9, 2009

clue444 said:
Find the relationship between a and b in the equation ax<sup>2</sup> + bx + c when the:

1) larger root exceeds the smaller root by 1

2) roots are in the ratio 2:3

can someone help me?

Do you have the answer?

Have you tried letting the roots be $\alpha$ and $\alpha + 1$ for the first one and for the second letting the roots be $2\alpha$ and $3\alpha$ ?

annabackwards · Jul 9, 2009

kurt.physics said:
Do you have the answer?

Have you tried letting the roots be $\alpha$ and $\alpha + 1$ for the first one and for the second letting the roots be $2\alpha$ and $3\alpha$ ?

Do what kurt says and use the sum and product of roots rules to find the relationships

Aquawhite · Jul 9, 2009

Not particularly too hard but very different to what I've come across. Thanks for posting ^_^

clue444 · Jul 9, 2009

kurt.physics said:
Do you have the answer?

Have you tried letting the roots be $\alpha$ and $\alpha + 1$ for the first one and for the second letting the roots be $2\alpha$ and $3\alpha$ ?

Yes I have tried that, but I am stuck with

-b/a = 2 $\alpha$ + 1

c/a = $\alpha$ ^2 + $\alpha$

what do I do now to get it into an a and b relationship??

(do not have the answer btw)

clue444 · Jul 9, 2009

Aquawhite said:
Not particularly too hard but very different to what I've come across. Thanks for posting ^_^

do you know how to do it? can u show me

?

tommykins · Jul 9, 2009

Try to rearrange to find $\alpha$ and sub it into the eqn.

clue444 · Jul 9, 2009

Hey guys I might be a bit slow, but I still haven't worked this out. I was wondering if anyone would be able to give the steps on doing this. I've done the below, not sure if I'm going the right way and don't know what to do now... thanks.

$\frac{-b}{a}=2\alpha +1$ --- equation 1

$\frac{c}{a}=\alpha^{2}+\alpha$ ---- equation 2

from equation 1 -> $-1-\frac{b}{a}=2\alpha$

$-\frac{1}{2}-\frac{b}{2a}=\alpha$ --- sub this into c/a

$\frac{c}{a}=(-\frac{1}{2}-\frac{b}{2a})^{2}+(-\frac{1}{2}-\frac{b}{2a})$

$= (\frac{-a-b}{2a})^{2}-\frac{1}{2}-\frac{b}{2a}$

$=\frac{a^2+2ab+b^2}{4a^2}-\frac{2a^2}{4a^2}-\frac{2ab}{4a^2}$

$\frac{c}{a}=\frac{b^2-a^2}{4a^2}$

Quadratic functions question. Need help! (1 Viewer)

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