# How do you solve by completing the square 3x^2-2x-3=0? and solve by quadratic formula 2x^2-4x+1=0 find the length and width of a rectangle whose perimeter is 16 units and whose area is 11 square units?

##### 1 Answer
Apr 3, 2015

No. 1
$3 {x}^{2} - 2 x - 3 = 0$

$3 \left({x}^{2} - \frac{2}{3} x - 1\right) = 0$

Now take the coefficient of $x$ and divide it by $2$ and square it,
$- \frac{2}{3} \to - \frac{1}{3} \to + \frac{1}{9}$

then add it and subtract it,

$3 \left({x}^{2} - \frac{2}{3} x + \frac{1}{9} - \frac{1}{9} - 1\right) = 0$

Since ${x}^{2} - \frac{2}{3} x + \frac{1}{9} = {\left(x - \frac{1}{3}\right)}^{2}$

$\implies 3 \left[{\left(x - \frac{1}{3}\right)}^{2} - \frac{1}{9} - 1\right] = 0$

$\implies {\left(x - \frac{1}{3}\right)}^{2} - \frac{10}{9} = 0$

$\implies \left(x - \frac{1}{3} - \sqrt{\frac{10}{9}}\right) \left(x - \frac{1}{3} + \sqrt{\frac{10}{9}}\right) = 0$

$\implies x - \frac{1}{3} - \sqrt{\frac{10}{9}} = 0 \implies x = \frac{1 + \sqrt{10}}{3}$

$\implies x - \frac{1}{3} + \sqrt{\frac{10}{9}} = 0 \implies x = \frac{1 - \sqrt{10}}{3}$

No. 2
$2 {x}^{2} - 4 x + 1 = 0$

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

$\implies x = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2 \sqrt{2}}{4} = 2 \pm \frac{\sqrt{2}}{2}$

No. 3
Perimeter : $2 \left(L + W\right) = 16$ ... (1)

Area : $L \times W = 11$ ... (2)

Solve the two equations simultaneously,

In (1), $L + W = 8 \implies L = 8 - W$

sub $L$ in (2)

$\implies \left(8 - W\right) \cdot W = 11 \implies {W}^{2} - 8 W + 11 = 0$

$\implies W = \frac{8 \pm \sqrt{64 - 44}}{2} \implies W = 4 \pm \sqrt{5}$

Back to equation (1), put $W$ in (1),

$\implies L = 8 - W \implies L = 8 - 4 \pm \sqrt{5} \implies L = 4 \pm \sqrt{5}$

Note : when $W = 4 - \sqrt{5}$ , $L = 4 + \sqrt{5}$