$$x^{ 2 }-2x-15=0$$
By factoring, I get:
$$(x-5)(x+3)$$
Which has the solutions:
$$x=5, x=-3$$
However when I use the quadratic formula (which is what the book saids to use), I get
$$\frac { 2 \pm \sqrt { 4-(4\cdot1\cdot(-15)) } }{ -2 } =$$ $$\frac { 2\pm 8 }{ -2 } $$
Which I evaluate to be $$x=-5, x=3$$
Where am I going wrong?
$\endgroup$ 23 Answers
$\begingroup$$$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$\implies x^2-2x-15=0\implies x=\frac{2\pm\sqrt{(-2)^2-4\cdot1(-15)}}{2\cdot1}=\frac{2\pm8}2=5,-3$$
$\endgroup$ 3 $\begingroup$Your error is that the denominator should be $2$, not $-2$
$\endgroup$ $\begingroup$The quadratic formula is $$ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$ The denominator should be $2$ because \[ a=1 \] \[ b=-2 \] \[ c=-15 \] So then \[ x=\frac{2\pm \sqrt{4-4(-15)}}{2}= \frac{2\pm \sqrt{64}}{2}= \frac{2\pm 8}{2}=1\pm 4 \] Thus $$ x=5 $$ $$ x=-3 $$
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