Question

One of the x-intercepts of the parabola represented by the equation y = 3x^2 + 6x − 10 is approximately (1.08, 0).

Answer 1

Answer:

$(-3.08,0)$.

Step-by-step explanation:

We have been given that one of the x-intercepts of the parabola $y=3x^2+6x-10$ is approximately (1.08, 0). We are asked to find the another x-intercept of parabola.

We will use quadratic formula to solve our given problem.

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

Upon substituting our given values in above formula we will get,

$x=\frac{-6\pm \sqrt{6^2-4*3*-10}}{2*3}$

$x=\frac{-6\pm \sqrt{36+120}}{6}$  

$x=\frac{-6\pm \sqrt{156}}{6}$    

$x=\frac{-6+\sqrt{156}}{6}\text{ or }x=\frac{-6-\sqrt{156}}{6}$

$x=-1+2.081665999466\text{ or }x=-1-2.081665999466$

$x=1.081665999466\text{ or }x=-3.081665999466$

$x\approx 1.08\text{ or }x\approx -3.08$

Therefore, the other x-intercept of the parabola is approximately $(-3.08,0)$.

Answer 2

3x^2 + 6x - 10 = 0

x  = [-6 +/- sqrt(^2 - 3*3*-10)] / 2*3

=  1.08 and -3.08

Other x -intercept is (-3.08,0)