Question

Match the following quadratic equations with their respective roots.

• A. $-\frac{4}{3}$, 3
• B. $-\frac{1}{3}$, 2
• C. 2,2

Answer 1

1) Given equation is $-3x^2+5x+12=0$
On comparing with $a{x}^2+bx+c=0$, we get
a=-3, b=5 and c=12
By quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-\left(5\right)\pm \sqrt{(5{)}^2-4(-3\left)\right(12)}}{2(-3)}=\frac{-5\pm \sqrt{25+144}}{-6}=\frac{-5\pm \sqrt{169}}{-6}=\frac{-5\pm 13}{-6}=\frac{8}{-6},\frac{-18}{-6}=-\frac{4}{3},3$

So, $-\frac{4}{3}$ and 3 are two roots of the given equation.

2) From the given quadratic equation,
a = 3, b = -5, c = -2

Using the quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2-4(3\left)\right(-2)}}{2\left(3\right)}$
$x=\frac{5\pm \sqrt{25+24}}{6}$
$x=\frac{5\pm 7}{6}$
$x=\frac{5+7}{6},\frac{5-7}{6}$
$x=\frac{12}{6},\frac{-2}{6}$
$x=2,\frac{-1}{3}$

3) $(x-2{)}^2=0$
so, x-2=0, x-2=0
or x=2, 2