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Vikki [24]
2 years ago
5

What are the zeros of the quadratic function? y = x^2 - 2x - 15

Mathematics
1 answer:
Pie2 years ago
3 0

Answer:

x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}

Step-by-step explanation:

-> 0=2x^2-2x-15

-zeros means the solution aka x intercept, meaning y=0

x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:2\left(-15\right)}}{2\cdot \:2}

=> according to quadratic formula

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

where ax^2+bx+c=0

=> x_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{31}}{2\cdot \:2}

simplify

x_1=\frac{-\left(-2\right)+2\sqrt{31}}{2\cdot \:2},\:x_2=\frac{-\left(-2\right)-2\sqrt{31}}{2\cdot \:2}

x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}

Cannot simplify farther

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