one x-int for a parabola is at the point (4,0) what is the x-int for the parabola defined by this equation y=x^2-x-13

Question

campbell_st · Answer

well it seems that the sero's have been rounded to the nearest whole number

so use the general quadratic formula to find the other zero then round the answer

plainntall · Answer

I don't understand the information provided.  You sure there aren't any typos?

The point (4,0) is not on the parabola give by the equation
y=x^2-x-13 as you can verify by substituting x=4
y=(4)^2-(4)-13 =-1 not 0 as you would expect for an x intercept.

If the equation is y=x^2-x-12 then that is a different story.

campbell_st · Answer

using the general quadratic formula you have 

$$x = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 1 \times (-13)}}{2}$$

which results in 

$$x = \frac{1 \pm \sqrt{53}}{2}$$

so if you look at this 

$$x = \frac{1 + \sqrt{53}}{2} \approx 4$$

the root has been rounded to perhaps the nearest whole number. 

Evaluate the the other root to get the answer

$$x = \frac{1 - \sqrt{53}}{2}$$