The area of a rectangular playing field is 192square meters. The length of the field is x+12 and the width is x-4. How do you calculate x by using quadratic formula?

Jul 14, 2017

$x = 12$

Explanation:

We know that the area formula for a rectangle is:

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So, we can plug these numbers in and then write everything in terms of a quadratic which we can solve with the quadratic formula.

$\left(x + 12\right) \times \left(x - 4\right) = 192$

Let's use the FOIL method to expand the left side.

${\underbrace{\left(x\right) \left(x\right)}}_{\text{First" + underbrace((x)(-4)) _ "Outer" + underbrace((12)(x)) _ "Inner" + underbrace((12)(-4))_"Last}} = 192$

${x}^{2} + \left(- 4 x\right) + \left(12 x\right) + \left(- 48\right) = 192$

${x}^{2} + 8 x - 48 = 192$

Now subtract $192$ from both sides.

${x}^{2} + 8 x - 240 = 0$

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This is a quadratic, so we can use the quadratic formula to solve it.

$a = 1$
$b = 8$
$c = - 240$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Now plug in all of those values and simplify.

$x = \frac{- \left(8\right) \pm \sqrt{{\left(8\right)}^{2} - 4 \left(1\right) \left(- 240\right)}}{2 \left(1\right)}$

$x = \frac{- 8 \pm \sqrt{64 + 960}}{2}$

$x = \frac{- 8 \pm \sqrt{1024}}{2}$

Note that $1024 = {2}^{10} = {\left({2}^{5}\right)}^{2} = {32}^{2}$

$x = \frac{- 8 \pm \sqrt{{32}^{2}}}{2}$

$x = \frac{- 8 \pm 32}{2}$

$x = - 4 \pm 16$

This means our two values of $x$ are:

$x = - 4 - 16 \text{ " and " } x = - 4 + 16$

$x = - 20 \text{ " and " } x = 12$

Remember that $x$ represents a length, and so it cannot possibly be negative. This leaves us with only one solution:

$x = 12$