Question

If the difference of the roots of a quadratic equation is 4 and the difference of their cubes is 208, then the quadratic equation is $x^2\pm 8x+12=0$
State true or false.

• A. True
• B. False

Answer 1

The correct option is A True

Let the roots of the equation be a and b
then, $a^3-b^3=208$
and $a-b=4$
cubing both sides:
$(a-b{)}^3=64$
$a^3-b^3-3ab(a-b)=64$
$208-3ab\left(4\right)=64$
$144=12ab$
$ab=12$
Similarly, $(a+b{)}^2=(a-b{)}^2+4ab$
$(a+b{)}^2=4^2+4(12)$
$(a+b{)}^2=16+48$
$a+b=\pm 8$
The general form of equation is $x^2-Sx+P=0$, hence the equation will be
$x^2\pm 8x+12=0$