How do I use the quadratic formula to solve $3 x^{2} - 6 = 4 x$ $3x^2 - 6 = 4x$ ?

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How do I use the quadratic formula to solve $3 x^{2} - 6 = 4 x$ $3x^2 - 6 = 4x$ ?

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Answer 1 · 2018-03-21T07:53:27

$x = \frac{2}{3} \pm \frac{\sqrt{22}}{3}$ $x = 2/3+-sqrt(22)/3$

Explanation:

Given:

$3 x^{2} - 6 = 4 x$ $3x^2-6 = 4x$

Subtract $4 x$ $4x$ from both sides to get:

$3 x^{2} - 4 x - 6 = 0$ $3x^2-4x-6 = 0$

This is now in standard form:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

with $a = 3$ $a=3$ , $b = - 4$ $b=-4$ and $c = - 6$ $c=-6$

So we can apply the quadratic formula to find:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (3) (- 6)}}{2 (3)}$ $color(white)(x) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(3))(color(blue)(-6))))/(2(color(blue)(3)))$

$x = \frac{4 \pm \sqrt{16 + 72}}{6}$ $color(white)(x) = (4+-sqrt(16+72))/6$

$x = \frac{4 \pm \sqrt{88}}{6}$ $color(white)(x) = (4+-sqrt(88))/6$

$x = \frac{4 \pm \sqrt{2^{2} \cdot 22}}{6}$ $color(white)(x) = (4+-sqrt(2^2 * 22))/6$

$x = \frac{4 \pm 2 \sqrt{22}}{6}$ $color(white)(x) = (4+-2sqrt(22))/6$

$x = \frac{2 \pm \sqrt{22}}{3}$ $color(white)(x) = (2+-sqrt(22))/3$

$x = \frac{2}{3} \pm \frac{\sqrt{22}}{3}$ $color(white)(x) = 2/3+-sqrt(22)/3$

Bonus

Notice in the derivation above that we had a factor $2^{2}$ $2^2$ in the radicand that we brought outside the square root as $2$ $2$ and then cancelled with the $2$ $2$ in the divisor.

This always happens when $b$ $b$ is even, adding an extra simplification step or two.

In such cases we can instead think of the original quadratic as taking the form:

$a x^{2} + 2 d x + c = 0$ $ax^2+2dx+c = 0$

Then the solutions can be expressed as:

$x = \frac{- d \pm \sqrt{d^{2} - a c}}{a}$ $x = (-d+-sqrt(d^2-ac))/a$

In our example $a = 3$ $a=3$ , $d = - 2$ $d=-2$ and $c = - 6$ $c=-6$ , so we get:

$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - (3) (- 6)}}{3}$ $x = (-(color(blue)(-2))+-sqrt((color(blue)(-2))^2-(color(blue)(3))(color(blue)(-6))))/(color(blue)(3))$

$x = \frac{2 \pm \sqrt{4 + 18}}{3}$ $color(white)(x) = (2+-sqrt(4+18))/3$

$x = \frac{2 \pm \sqrt{22}}{3}$ $color(white)(x) = (2+-sqrt(22))/3$

$x = \frac{2}{3} \pm \frac{\sqrt{22}}{3}$ $color(white)(x) = 2/3+-sqrt(22)/3$

I don't know how widely this variant of the quadratic formula is used.

If $a = 1$ $a=1$ then we can simplify further to say that the solutions of:

$x^{2} + 2 d x + c = 0$ $x^2+2dx+c = 0$

are:

$x = - d \pm \sqrt{d^{2} - 4 c}$ $x = -d+-sqrt(d^2-4c)$

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How do I use the quadratic formula to solve $3 x^{2} - 6 = 4 x$ $3x^2 - 6 = 4x$ ?

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How do I use the quadratic formula to solve $3 x^{2} - 6 = 4 x$ $3x^2 - 6 = 4x$ ?