$\displaystyle x^2+p=4x+2$ is three time the other roots.Find the value of $\displaystyle p$.
please help..i dont know how to do..
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 $\displaystyle \sqrt{16-4(p-2)}=2$
$\displaystyle 16-4(p-2)=4$
$\displaystyle 12=4(p-2)$
$\displaystyle 3=p-2$