How do you solve the quadratic using the quadratic formula given #a^2-7a-10=0#?

1 Answer
George C.
Aug 10, 2016

#a = 7/2+-sqrt(89)/2#

Explanation:

The slight nuisance with this question is that the #a# used as a variable name clashes with the #a, b, c# that normally stand for the coefficients of the terms of a quadratic.

To sidestep this issue, consider the quadratic equation:

#x^2-7x-10 = 0#

This has the same zeros as our original quadratic, just expressed as values of #x# instead of values of #a#.

Our quadratic equation in #x# is of the form #ax^2+bx+c = 0# with #a=1#, #b=-7# and #c=-10#.

It therefore has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (7+-sqrt((-7)^2-4(1)(-10)))/(2*1)#

#= (7+-sqrt(49+40))/2#

#= (7+-sqrt(89))/2#

#= 7/2+-sqrt(89)/2#

So the zeros of our original quadratic are given by:

#a = 7/2+-sqrt(89)/2#

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