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

Sagot :

Answer:

[tex]x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}[/tex]

Step-by-step explanation:

[tex]-> 0=2x^2-2x-15[/tex]

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

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

=> according to quadratic formula

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

where [tex]ax^2+bx+c=0[/tex]

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

simplify

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

[tex]x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}[/tex]

Cannot simplify farther

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