Question

Find b and then solve the equation

2x^2+bx-10=0 if one of its roots is 5

Answer 1

well, since we know that one root is 5, that is a factor of it is (x-5), so hmmm by the remainder theorem we also know that if we plug "5" as our argument in our f(x), that is, if we do f(5), it must result to 0. so f(5) = 0, if that's so then

`2x^2+bx-10=f(x)\implies 2(5)^2+b(5)-10=f(5) \\\\\\ 2(5)^2+b(5)-10=0\implies 50+5b-10=0\implies 5b=-40 \\\\\\ b=\cfrac{-40}{5}\implies \boxed{b=-8} \\\\[-0.35em] ~\dotfill\\\\ 2x^2-8x-10=0\implies 2(x^2-4x-5)=0\implies 2\stackrel{ {\Large \begin{array}{llll} x=-1 \end{array}} }{(x+1)}(x-5)=0`