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?

Question

2502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-03T19:54:01

**No. 1 **
$3x^2 - 2x - 3 = 0$

$3(x^2 - 2/3x -1) = 0$

Now take the coefficient of $x$ and divide it by $2$ and square it,
$-2/3 -> -1/3 -> +1/9$

then add it and subtract it,

$3(x^2 - 2/3x + 1/9 -1/9 - 1) = 0$

Since $x^2 -2/3x + 1/9 = (x - 1/3)^2$

$=> 3[(x - 1/3)^2 -1/9 -1] = 0$

$=> (x - 1/3)^2 - 10/9 = 0$

$=> (x - 1/3 - sqrt(10/9))(x - 1/3 + sqrt(10/9)) = 0$

$=> x - 1/3 - sqrt(10/9) = 0 => x = (1 + sqrt(10))/3$

$=> x - 1/3 + sqrt(10/9) = 0 => x = (1 - sqrt(10))/3$

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

$x = (-(-4) +- sqrt((-4)^2 -4*2*1))/(2*2)$

$=> x = (4 +- sqrt(8))/4 = (4 +- 2sqrt(2))/4 = 2 +-sqrt(2)/2$

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

Area : $LxxW = 11$ ... (2)

Solve the two equations simultaneously,

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

sub $L$ in (2)

$=> (8 - W)*W = 11 => W^2 -8W + 11 = 0$

$=> W = (8 +- sqrt(64 - 44))/2 => W = 4 +- sqrt(5)$

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

$=> L = 8 - W => L = 8 - 4 +-sqrt(5) => L = 4 +- sqrt(5)$

Note : when $W = 4 - sqrt(5)$ , $L = 4 + sqrt(5)$

How do you solve by completing the square 3x^2-2x-3=03x^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?