Sally A.

asked • 05/09/20

What is an example of a quadratic equation that has two solutions but cannot be solved by factoring

What is an example of a quadratic equation that has two solutions but cannot be solved by factoring. and how do you know it cannot be solved by factoring.

Brenda D.

tutor
Almost anything that has a constant term that cannot be factored in such a way to sum to the middle term and requires use of the quadratic equation. For example 275 + 4x - 4x^2; You can test a quadratic against FOIL if it does not unFOIL so to speak you have to use the quadratic equation or try completing the square or review the discriminate as described below
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05/09/20

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Jon S. answered • 05/09/20

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x^2-8x-15


discriminant = b^2 - 4 a c = 64 - 4(1)(-15) = 124 so there are two real solutions.


To solve by factoring would need two numbers than when multiplied = -15 and when summed = -8


Only numbers which multiplied equal -15 are

-1 and 15 (sum 14)

1 and -15 (sum -14)

3 and -5 (sum -2)

-3 and 5 (sum 2)


None of which add to -8


Have to solve using quadratic formula: 4 +/- sqrt(31)

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Sally A.

thank you!
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05/09/20

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