Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:
a)value of p
b) root of the quadratic equation..
anyone sir...please help!
Hi mastermin346,
sa-ri-ga-ma's method is a fast way...
the long way to do it is
the equation in $\displaystyle ax^2+bx+c$ form is $\displaystyle 8x^2-18x+(3p-4)=0$
One root is $\displaystyle \frac{18+\sqrt{18^2-32(3p-4)}}{16}$
The other root is $\displaystyle \frac{18-\sqrt{18^2-32(3p-4)}}{16}$
The larger root must be the first one if the roots are real.
Then twice the second root is the first.
$\displaystyle 2\left(18-\sqrt{324-32(3p-4)}\right)=18+\sqrt{324-32(3p-4)}$
$\displaystyle 18=3\sqrt{324-32(3p-4)}$
$\displaystyle \sqrt{324-32(3p-4)}=6$
$\displaystyle 324-32(3p-4)=36$
$\displaystyle 32(3p-4)=324-36=288$
$\displaystyle 4(3p-4)=36$
$\displaystyle 3p-4=9$
$\displaystyle p=\frac{13}{3}$
The equation is
$\displaystyle 8x^2-18x+13-4=0$
$\displaystyle 8x^2-18x+9=0$
Either find the roots of this or substitute "p" into
$\displaystyle \frac{18\pm\sqrt{324-32(9)}}{16}$