Question

Find the root of quadratic equation 6xx + x -15

Answer 1

The roots of the quadratic equation are $\frac{3}{2}$ and $-\frac{5}{3}$.

Answer 2

To find the roots of the quadratic equation $6x^2 + x - 15 = 0$, we use the quadratic formula.

A quadratic equation of the form $ax^2 + bx + c = 0$ has roots given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In the given equation, $6x^2 + x - 15 = 0$:

$a = 6$ $b = 1$ $c = -15$

Substitute these values into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(6)(-15)}}{2(6)}$ $x = \frac{-1 \pm \sqrt{1 - (-360)}}{12}$ $x = \frac{-1 \pm \sqrt{1 + 360}}{12}$ $x = \frac{-1 \pm \sqrt{361}}{12}$

Since $\sqrt{361} = 19$:

$x = \frac{-1 \pm 19}{12}$

This gives two possible roots:

1. $x_1 = \frac{-1 + 19}{12} = \frac{18}{12} = \frac{3}{2}$
2. $x_2 = \frac{-1 - 19}{12} = \frac{-20}{12} = \frac{-5}{3}$

The roots of the quadratic equation $6x^2 + x - 15 = 0$ are $x = \frac{3}{2}$ and $x = -\frac{5}{3}$.

Explanation:

Identify coefficients $a=6$, $b=1$, $c=-15$. Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Calculate the discriminant $\sqrt{1^2 - 4(6)(-15)} = \sqrt{1+360} = \sqrt{361} = 19$. Substitute back to find roots: $x = \frac{-1 \pm 19}{12}$, yielding $x = \frac{18}{12} = \frac{3}{2}$ and $x = \frac{-20}{12} = -\frac{5}{3}$.