Question

find the zeros of the quadratic equation x^2+3x+10
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Answer 1

Answer:

1. take x=0
2. 2+3(0)+10
3. 2+0+10
4. 12 is the final ans

Answer 2

The zeros of the quadratic equation x²+3x+10 are $\alpha =\frac{-3+\sqrt{31 } }{2}$ and $\beta =\frac{-3-\sqrt{31 } }{2}$

Explanation:

Given:

x²+3x+10

To find:

The zeros of the quadratic equation

Formula:

$\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

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

Solution:

==> x²+3x+10

==> a = coefficient of x²

==> b = coeffcient of x

==> c = constant

==> a = 1

==> b = 3

==> c =10

==> Apply these values in the formula

==> $\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

==> $\alpha =\frac{-3+\sqrt{3^{2}-4(1)(10) } }{2(1)}$

==>$\alpha =\frac{-3+\sqrt{9-4(10) } }{2}$

==> $\alpha =\frac{-3+\sqrt{9-40 } }{2}$

==> $\alpha =\frac{-3+\sqrt{31 } }{2}$

==> $\beta =\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

==> $\beta =\frac{-3-\sqrt{3^{2}-4(1)(10) } }{2(1)}$

==>$\beta =\frac{-3-\sqrt{9-4(10) } }{2}$

==> $\beta =\frac{-3-\sqrt{9-40 } }{2}$

==> $\beta =\frac{-3-\sqrt{31 } }{2}$

The zeros of the quadratic equation x²+3x+10 are $\alpha =\frac{-3+\sqrt{31 } }{2}$ and $\beta =\frac{-3-\sqrt{31 } }{2}$