How do you find the solution to the quadratic equation $x^{2} - 4 x - 3 = 0$ $x^2 - 4x -3 = 0$ ?

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How do you find the solution to the quadratic equation $x^{2} - 4 x - 3 = 0$ $x^2 - 4x -3 = 0$ ?

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Answer 1 · 2018-04-14T11:21:11

$x = 2 \pm \sqrt{7}$ $x=2+-sqrt7$

Explanation:

$there are no whole numbers which multiply to - 3$ $"there are no whole numbers which multiply to - 3"$
$and sum to - 4$ $"and sum to - 4"$

$we can solve using the method of completing the square$ $"we can solve using the method of "color(blue)"completing the square"$

$the coefficient of the x^{2} term is 1$ $"the coefficient of the "x^2" term is 1"$

$∙ add subtract {(\frac{1}{2} coefficient of the x-term)}^{2} to$ $• " add subtract "(1/2"coefficient of the x-term")^2" to"$
$x^{2} - 4 x$ $x^2-4x$

$\Rightarrow x^{2} + 2 (- 2) x + 4 - 4 - 3 = 0$ $rArrx^2+2(-2)xcolor(red)(+4)color(red)(-4)-3=0$

$\Rightarrow {(x - 2)}^{2} - 7 = 0$ $rArr(x-2)^2-7=0$

$\Rightarrow {(x - 2)}^{2} = 7$ $rArr(x-2)^2=7$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\Rightarrow x - 2 = \pm \sqrt{7} \leftarrow note plus or minus$ $rArrx-2=+-sqrt7larrcolor(blue)"note plus or minus"$

$\Rightarrow x = 2 \pm \sqrt{7} \leftarrow exact solutions$ $rArrx=2+-sqrt7larrcolor(red)"exact solutions"$

Answer 2 · 2018-04-14T11:26:45

x = $2 \pm \sqrt{7}$ $2+- sqrt(7)$

Explanation:

Apply quadratic formula for this equation instead of trying to factor it out.
1/ $⎛ ⎜ ⎜ ⎝ \frac{- b \pm \sqrt{{(b)}^{2} - 4 (a) (c)}}{2 (a)} ⎞ ⎟ ⎟ ⎠$ $((-b+-sqrt((b)^2-4(a)(c)))/(2(a)))$

2/ $⎛ ⎜ ⎜ ⎝ \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (- 3)}}{2 (1)} ⎞ ⎟ ⎟ ⎠$ $((-(-4)+-sqrt((-4)^2-4(1)(-3)))/(2(1)))$

3/ $(\frac{4 \pm \sqrt{16 + 12}}{2})$ $((4+-sqrt(16+12))/(2))$

4/ $(\frac{4 \pm 2 \sqrt{7}}{2})$ $((4+-2sqrt(7))/(2))$ ( 2 cancel out)

5/ x = $2 \pm \sqrt{7}$ $2+-sqrt(7)$

Answer 3 · 2018-04-14T11:28:07

$x = 2 + \sqrt{7} or x = 2 - \sqrt{7}$ $x=2+sqrt7 or x=2-sqrt7$

Explanation:

Here,

$x^{2} - 4 x - 3 = 0$ $x^2-4x-3=0$

$\Rightarrow x^{2} - 4 x + 4 - 7 = 0$ $=>x^2-4x+4-7=0$

$\Rightarrow {(x - 2)}^{2} = 7 = {(\sqrt{7})}^{2}$ $=>(x-2)^2=7=(sqrt7)^2$

$\Rightarrow x - 2 = \pm \sqrt{7}$ $=>x-2=+-sqrt7$

$\Rightarrow x = 2 \pm \sqrt{7}$ $=>x=2+-sqrt7$

OR

Comparing with quadratic equation,

$a x^{2} + b x + c = 0 \Rightarrow a = 1, b = - 4, c = - 3$ $ax^2+bx+c=0=>a=1,b=-4,c=-3$

$△ = b^{2} - 4 a c = {(- 4)}^{2} - 4 (1) (- 3)$ $triangle=b^2-4ac=(-4)^2-4(1)(-3)$

$\Rightarrow △ = 16 + 12 = 28 = 4 \times 7$ $=>triangle=16+12=28=4xx7$

$\sqrt{△} = 2 \sqrt{7}$ $sqrt(triangle)=2sqrt7$

So,

$x = \frac{- b \pm \sqrt{△}}{2 a}$ $x=(-b+-sqrt(triangle))/(2a)$

$x = \frac{4 \pm 2 \sqrt{7}}{2 (1)}$ $x=(4+-2sqrt7)/(2(1))$

$x = 2 \pm \sqrt{7}$ $x=2+-sqrt7$

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