\(\displaystyle x^2+p=4x+2\) One root is three times the other root. Find the value of \(\displaystyle p\).

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

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

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

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

\(\displaystyle 3=p-2\)