$\displaystyle x^2+p=4x+2$ is three time the other roots.Find the value of $\displaystyle p$.

please help..i dont know how to do..(Crying)

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- May 24th 2010, 09:18 AMmastermin346quadratic equation problem
$\displaystyle x^2+p=4x+2$ is three time the other roots.Find the value of $\displaystyle p$.

please help..i dont know how to do..(Crying) - May 24th 2010, 09:33 AMDinkydoe
Can you be a little more specific?

Do you mean that if $\displaystyle x=a$ is a root, then the other root is $\displaystyle x=3a$?

Then find the corresponding $\displaystyle p$? - May 24th 2010, 09:38 AMmastermin346
- May 24th 2010, 10:09 AMpapex
in that case, rewriting the equation as: x^2-4x+p-2=0 we have -4=4a and p-2=3a^2

thus a=-1 and p=5

the equation becomes x^2-4x+3=0 which has 1 and 3 as roots - May 24th 2010, 10:12 AMDinkydoe
Ok, write the equation as $\displaystyle x^2-4x+(p-2)= 0$

Suppose that $\displaystyle x^2-4x+(p-2) = (x-a)(x-3a)= x^2-4ax+3a^2$

Now we have $\displaystyle a=1, 3a=3$ and we can solve $\displaystyle p-2=3a^2$. - May 24th 2010, 10:28 AMArchie Meade
Hi mastermin346,

Factorising is the fastest way.

$\displaystyle (x-root_1)(x-root_2)=0,\ x(x-root_2)-root_1(x-root_2)=0,$

$\displaystyle x^2-(sum\ of\ roots)x+product\ of\ roots=0$

Here the sum of the roots=4.

One is 3 times the other, so the sum of the roots is 4 times the smaller root.

Smaller root = 4/4.

If you had an expression that was difficult to factorise,

you could proceed as follows.....

The solutions (roots) of a quadratic equation $\displaystyle ax^2+bx+c=0$

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

One root is $\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}$

The other root (a quadratic has at most 2 roots) is $\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}$

First re-arrange your equation to $\displaystyle ax^2+bx+c=0$

$\displaystyle x^2+p=4x+2\ \Rightarrow\ (1)x^2+(-4)x+(p-2)=0$

$\displaystyle a=1$

$\displaystyle b=-4$

$\displaystyle c=p-2$

If one root is 3 times the other, the bigger root__must__be the one with the +

$\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}=3\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$

The denominator is common, so it is irrelevant.

$\displaystyle -b+\sqrt{b^2-4ac}=3\left(-b-\sqrt{b^2-4ac}\right)$

$\displaystyle 4+\sqrt{16-4(p-2)}=3\left(4-\sqrt{16-(p-2)}\right)$

$\displaystyle \sqrt{16-4(p-2)}+3\sqrt{16-4(p-2)}=12-4=8$

$\displaystyle 4\sqrt{16-4(p-2)}=8$

This gives $\displaystyle \sqrt{16-4(p-2)}=2$

$\displaystyle 16-4(p-2)=4$

$\displaystyle 12=4(p-2)$

$\displaystyle 3=p-2$