How can you find $m$ in $mx^2+(m-3)x+1=0 $ so that there is only one solution

Question

How can you find $m$ in $$mx^2+(m-3)x+1=0 $$ so that there is only one solution.

I tried to solve it by quadratic equation but I end up with two solutions.

So I want it know that is there a way so that I'll only get one solution the end.

Thanks.

Answer 1

A quadratic equation always has $2$ solutions, by the Fundamental Theorem of Algebra.

So if it has one distinct solution, then the two roots must be same.

Hence the discriminant must be equated to $0$.

Therefore, $$(m-3)^2-4\cdot m\cdot 1=0$$ $$m^2-10m+9=0$$ $$(m-9)(m-1)=0$$

So required value of $m$ is $1$ and $9$.

Answer 2

If we want only one solution, we need the discriminant of the quadratic equation to equal $0$. That is, in $$x = {3 - m \pm \sqrt{(m-3)^2 - 4m} \over 2m},$$ we need the stuff under the radical $(m-3)^2 - 4m = 0$. So $$m^2 - 6m + 9 - 4m = m^2 - 10m + 9 = (m-9)(m-1)=0.$$ So we get two answers, $m = 9$ and $m=1$. These are the only two solutions because if this is the case, then $$x = {3-9\pm \sqrt{6^2-4(9)} \over 2(9)} = {-6 \pm 0 \over 18} = -{1 \over 3}$$ and $$x = {2 - 1 \pm \sqrt{(-2)^2-4} \over 2} = {1 \pm 0 \over 2} = {1 \over 2}$$

Notice how each case of $m=9$ and $m=1$ respectively yields only one solution.

Answer 3

Let $x=\alpha$ be one solution of the given quadratic equation: $mx^2+(m-3)x+1=0$ then $\alpha$ & $\alpha$ be the roots of the given equation $$\text{sum of roots}=\alpha+\alpha=-\frac{m-3}{m}$$$$\alpha=\frac{3-m}{2m}\tag 1$$ $$\text{product of roots}=\alpha\cdot \alpha=\frac{1}{m}$$$$\alpha^2=\frac{1}{m}\tag 2$$ substituting the value of $\alpha$ from (2) into (1), one should get $$\left(\frac{3-m}{2m}\right)^2=\frac 1m$$ $$m^2-10m+9=0$$ using quadratic formula, $$m=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(9)}}{2(1)}=5\pm4$$ Thus, the given quadratic equation: $mx^2+(m-3)x+1=0$ will have only one solution for $\color{red}{m=9}\ \text{or} $ $ \ \ \color{red}{m=1}$.