How do you solve #36x = —4x^2 — 50#?

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Answer 1 · 2018-04-06T19:46:14

The set of solutions is #\{\frac{\sqrt{31}-9}{2}, \frac{-9-\sqrt{31}}{2}\}#.

#36x=-4x^2-50#

#\implies 4x^2+36x+50=0#

Applying the quadratic formula, which states that if #ax^2+bx+c=0#, then:

#x=\frac{-b\pm\sqrt{\Delta}}{2a}#

Where #\Delta=b^2-4ac#. Solving the equation with our values:

#x_{12}=\frac{-36\pm\sqrt{36^2-4\cdot 4\cdot 50}}{8}#

#x_{12}=\frac{-36\pm 4\sqrt{31}}{8}#

#x_{12}=\frac{-9\pm \sqrt{31}}{2}#

And so the two solutions are #\frac{\sqrt{31}-9}{2}# and #\frac{-9-\sqrt{31}}{2}#.