I am revising for a mathematics exam and am looking over simultaneous equations. I was curious as to when I use the quadratic formula and when I don't? I realize there are multiple ways to solve a question - for example 2x^2+7x-15=0 - can I use the simultaneous equation for this or do I just use the factorizing method?
When do you use the quadratic formula?
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If you can factorize your quadratic without using the formula then you should do it, because it is usually faster.
When you have a quadratic $ax^2+bx+c$ you can easily factorize it if you can find two numbers $n_1,n_2$ such that $n_1+n_2=b$ and $n_1n_2=ac$ by rewriting $bx$ as $n_1x+n_2x$ and then grouping similar terms.
In the case of $2x^2+7x-15=0$ we have $n_1=10$ and $n_2=-3$ so we factorize it without having to use the quadratic formula:
$2x^2+7x-15=0$
$2x^2+10x-3x-15=0$
$2x(x+5)-3(x+5)=0$
$(2x-3)(x+5)=0$
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Key Idea $\ $ Use AC-method to reduce to factoring a polynomial that's $\,\rm\color{#c00}{monic}\,$ (lead coeff $=\color{#c00}1)$
$$\quad\ \ \begin{eqnarray} f &\,=\,& \ \ 2\ x^2+\ 7\ x\ -\,\ 15\\ \Rightarrow\ 2f &\,=\,&\ (2x)^2\! +7(2x)-30\\ &\,=\,& \ \ \ \color{#c00}{X^2+\,7\ X\ -\,\ 30},\,\ \ X\, =\, 2x\\ &\,=\,& \ \ \,(X-3)\ (X+\,10)\\ &\,=\,& \ \ (2x-3)\,(2x+10)\\ \Rightarrow\ f\,=\, 2^{-1}(2f) &\,=\,& \ \ (2x - 3)\,(x+5)\\ \end{eqnarray}\qquad\qquad$$
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It's my view that the quadratic formula should always be used to factor a quadratic. That's what it's for.
Completing the square should be used to transform a quadratic into vertex-focus form (and to derive the quadratic formula).
Factoring by grouping is only really taught at the HS level to give students an introduction to Number Theory (finding integer solutions to equations, analyzing factors and residuals, etc), or perhaps as a preliminary for developing strategies for factoring cubics. Other than that, it's not very useful.
I always try to use the factorizing method first. But sometimes the method does not work for certain quadratic equations. In that case, I resort to using the quadratic formula $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ For your equation of $2x^2+7x-15=0$, we have $a=2,b=7,c=-15$.