Values of m for which the expression is positive

Question

Q: Find the values of m for which the expression below is always positive. $x^2 + 2mx + (3m-2)$

I have attempted the question and know that I'm supposed to use the discriminant, however I'm having a bit of trouble with the substitution. I factorised first to get $4(m^2x^2-3m+2)$ then tried factorisation by grouping but I couldnt get it. Any ideas?

$$\begin{align}x^2+2mx+(3m-2)\\b^2-4ac>0\\(2mx)^2-4*1*(3m-2)\\4m^2x^2-12m+8=0\\4(m^2x^2-3m+2)=0\end{align}$$ ??

Thanks for your help! :)

Answer 1

Since $$4(m^2x^2 - 3m + 2)= 0$$ then because $0=0\times 4$, we can actually write that $$m^2x^2 - 3m + 2 = 0.$$ And since this equation is a quadratic equation (particularly a quadratic trinomial) with only one variable we want to solve for, namely $m$ (here, $m$ is known is the indeterminate) then we can use the quadratic formula to solve for $m$.

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Quadratic Fromula: Provided the equation $ax^2 + bx + c = 0$ for constants $a, b, c, x$ with $a\neq 0$, $$x=\frac{-b\pm\sqrt{\Delta}}{2}$$ such that $\Delta = b^2 - 4ac =$ the discriminant. Proofs of this formula can be found here.

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Now simply substitute.

Answer 2

Hint: If you want to represent $x^2 + 2mx + (3m-2)$ as $ax^2+bx+c$, then:

$$a=1, b=2m, c=3m-2$$

so the discriminant is:

$$b^2-4ac=(2m)^2-4\cdot 1\cdot (3m-2)$$

Hope you can take it from here.

Answer 3

$$x^2 + 2mx + (3m-2)=(x+m)^2-(m^2-3m+2)$$

Since $(x+m)^2 \ge 0$, all you need is to solve $$m^2-3m+2 < 0$$