How Many Solutions?

Algebra Level 2

For real numbers $a, b,$ and $c$ such that $a \ne \frac{c-b}{2},$ how many real values of $x$ satisfy the equation

$(x-a)(x-c)=(x+b)(a-x)?$

0 1 2 Insufficient information

1 solution

Michael Fuller
Jan 5, 2016

$(x-a)(x-c)=(x+b)\color{#D61F06}{(a-x)} ~\Rightarrow~ (x-a)(x-c)=\color{#D61F06}{\mathbf{-}(x-a)}(x+b)$ $\Rightarrow~ (x-a)(x-c)+(x-a)(x+b)=0 ~\Rightarrow~ (x-a)(2x-c+b)=0$

Therefore by the Zero Product Property we have $x-a=0 ~\Rightarrow~ \color{#3D99F6}{x=a}$ $2x-c+b=0 ~\Rightarrow~ \color{#3D99F6}{x=\dfrac{c-b}{2}}$

It is given that $a \ne \dfrac{c-b}{2}$ , so our two solutions are distinct.

Therefore there are $\large \color{#20A900}{\boxed{2}}$ real values of $x$ that satisfy the equation.

It is not correct to take x=a, because if we put x=a in (a−a)(x−c)=(x+b)(a−a) we get 0 (a-c) = 0 (a+b) and hence LHS and RHS cannot be compared.

Apoorv Panse - 1 year, 8 months ago

ok.but if we cancel (x-a) from two side then.......

sayak dutta - 3 years, 8 months ago

Can we do that? I thought we cannot divide both sides of any equation by an unknown since that may remove possible solutions. Doing so also sometimes creates wrong statements. For example, consider 2x = 3x This equation only has 1 real solution, x = 0 However, if you divide both sides by x: 2 = 3 you get a false statement!

Irfan Syahril - 9 months, 3 weeks ago

How Many Solutions?

1 solution

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