How do you write a quadratic equation in standard form that has two solutions, 6, and -10 with the leading coefficient one?

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How do you write a quadratic equation in standard form that has two solutions, 6, and -10 with the leading coefficient one?

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Answer 1 · 2016-09-29T05:06:39

$y = x^{2} + 4 x - 60$ $y=x^2+4x-60$

Explanation:

We want an equation which equals $0$ $0$ at the given points $6$ $6$ and $- 10$ $-10$ .

Our quadratic equation should be a product of expressions which are zero at the specified roots.
Consider $(x - 6) \cdot (x + 10) = 0$ $(x-6)*(x+10) = 0$

This equality holds if $x = 6$ $x=6$ since
$(6 - 6) \cdot (6 + 10) = 0 \cdot 16 = 0$ $(6-6)*(6+10) = 0*16 = 0$
And the equality holds if $x = - 10$ $x=-10$ since
$(- 10 - 6) \cdot (- 10 + 10) = - 16 \cdot 0 = 0$ $(-10-6)*(-10+10) = -16*0 = 0$

Expanding this equation by the FOIL method, we get:

$x^{2} + 10 x - 6 x - 60$ $x^2 + 10x - 6x - 60$

Combining like terms, we find our solution:

$x^{2} + 4 x - 60$ $x^2 +4x - 60$

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How do you write a quadratic equation in standard form that has two solutions, 6, and -10 with the leading coefficient one?

1 Answer

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