Question

One of the real roots of the quadratic equation $3x^2-5x-2=0$ is :

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

Answer 1

The correct option is C

2

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

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

or

The given quadratic equation can be factorised as :
$3x^2-5x-2=0$
$3x^2-6x+x-2=0$
$3x(x-2)+1(x-2)=0$
$(3x+1)(x-2)=0$
So, $x=2,\frac{-1}{3}$