How do you solve $x^{2} + 4 x - 21 = 0$ $x^2 + 4x - 21 = 0$ ?

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Answer 1 · 2015-06-01T20:07:43

We are going to solve it with $Delta$

The equation is of the form $ax²+bx+c=0$

$a=1$ $b=4$ $c=-21$

$Delta=b²-4ac$
$Delta=4²-4*1*(-21)$
$Delta=16+84$
$Delta=100$ ( $=sqrt10$ )

$x_1=(-b-sqrtDelta)/(2a)$

$x_1=(-4-10)/2$

$x_1=-14/2$

$x_1=-7$

$x_2=(-b+sqrtDelta)/(2a)$

$x_2=(-4+10)/2$

$x_2=6/2$

$x_2=3$

Answer 2 · 2015-06-01T20:17:46

You can solve this by using the general formula for complete quadratic equations.

$x=(-b+-sqrt(b^2-4ac))/(2a)$
For a complete quadratic equation $ax^2+bx+c=0$ .
Let's apply it to your equation, which gives:

$x=(-4+-sqrt(16-4*1*(-21)))/2=(-4+-sqrt(100))/2=(-4+-10)/2=-2+-5$
So, we've got two solutions:
$x=3$
and
$x=-7$

Answer 3 · 2015-06-01T20:51:54

A less direct, but perhaps simpler approach is to recognise that if $x^2+4x-21$ has linear factors, then they must look like:

$(x+p)(x-q) = x^2+(p-q)x-(pxxq)$ where $p, q > 0$

The reason I say it must look like this with one $+$ and one $-$ is because the sign of the constant term in the original quadratic is negative.

Then $p-q = 4$ and $pxxq = 21$ .

What pair of factors of $21$ differs by $4$ ?

There are not many choices of factorizations, basically either $1xx21$ or $3xx7$ .

$7-3 = 4$ and $7xx3 = 21$ , so let $p = 7$ and $q=3$

$x^2+4x-21 = (x+p)(x-q) = (x+7)(x-3)$

This is zero when $x=-7$ or $x=3$

How do you solve x^2 + 4x - 21 = 0x2+4x−21=0x^2 + 4x - 21 = 0?