one x-intercept for a parabola is at the point (3.22,0) find the other x-intercept for the parabola defined by this equation y=2x^2-8x+5 round to the nearest hundredth if necessary

Respuesta :

Answer:

(0.78,0)

Step-by-step explanation:

I would use the quadratic formula.

[tex]a=3[/tex]

[tex]b=-8[/tex]

[tex]c=5[/tex]

[tex]\text{ The quadratic formula is } x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{ Let's find } b^2-4ac \text{ first}\\(-8)^2-4(2)(5)\\64-8(5)\\64-40\\24\\\\\text{ Now let's find } -b\\-b=8\\\text{ And } 2a\\2(2)=4\\\\\text{ Let's plug in this information }\\x=\frac{8 \pm \sqrt{24}}{4}\\\\\text{ We are now going to simplify }\\x=\frac{8}{4} \pm \frac{\sqrt{24}}{4} \\x=2 \pm \frac{\sqrt{4 \cdot 6}}{4}\\x=2 \pm \frac{\sqrt{4} \sqrt{6}}{4} \\x=2 \pm \frac{2 \sqrt{6}}{4}\\[/tex]

[tex] x=2 \pm \frac{\sqrt{6}}{2}[/tex]

So let's put both of these into out calculator

2 + sqrt(6)/2                  and             2-sqrt(6)/2

One of them should be approximately 3.22 as the question suggests.

3.22                                                        0.78

So the other x-intercept is approximately (0.78,0)