For what values for parameter b will x have real results? $\frac{2x}{x+3}+\frac{bx-2x}{x^2-9}=\frac{x-1}{x-3}$

Question

For what values of parameter b will x have real results? $$ \frac{2x}{x+3}+\frac{bx-2x}{x^2-9}=\frac{x-1}{x-3}$$ I get to the point where I have: $x^2-(10-b)x+3=0$. If I need real results for x then D needs to be >0 but when I do that I get $b^2-20b+88>0$. What am I supposed to do with that? There is no way I can get b like that.

Answer 1

It should be $$b^2-20b+88\geq0.$$ Also, you need to check $$3^2+(b-10)3+3=0$$ and $$(-3)^2+(b-10)(-3)+3=0.$$ In these cases we obtain:

1. $b=6$, which gives $x^2-4x+3=0$, which gives a root $1$;

2. $b=14$, which gives $x^2+4x+3=0$ and a root $-1$.

Thus, $b=6$ and $b=14$ they are valid.

Answer 2

For

$$x^2-(10-b)x+3=0$$

to have real solutions we need

$$\Delta=b^2-20b+88\color{red}{\ge} 0$$

from here we can find the range for $b$ by the roots of the quadratic equation.

Note that for the solution of the original equation you also need to set $x\neq \pm 3$.

Answer 3

If you already know that $b^2-20b+88\ge 0$, then go on: $(b-10)^2-100+88\ge 0$, then $(b-10)^2\ge 12$. Now you surely can finish it.

Answer 4

$b^2-20b+88\ge0$

Solving the related equation $b^2-20b+88=0$ using the quadratic formula, $b=10\pm2\sqrt{3}$.

So, $(b-(10+2\sqrt{3}))(b-(10-2\sqrt{3}))\ge0$, which means $b\ge10+2\sqrt{3}$ or $b\le10-2\sqrt3$

Also, $x^2-10x+bx+3=0 \rightarrow bx=-x^2+10x-3\rightarrow b=\cfrac{-x^2+10x-3}{x}$ where $x\ne0$.