How to solve 2x-1/3 = x^2+2x/5 using the quadratic formula?

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How to solve 2x-1/3 = x^2+2x/5 using the quadratic formula?

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Answer 1 · 2016-01-17T04:43:18

$x = \frac{12 \pm \sqrt{69}}{15}$ $x=(12+-sqrt69)/15$

Explanation:

I am interpreting the given equation as:

$2 x - \frac{1}{3} = x^{2} + \frac{2 x}{5}$ $2x-1/3=x^2+(2x)/5$

Move these all to the same side of the equation.

$0 = x^{2} + \frac{2 x}{5} - 2 x + \frac{1}{3}$ $0=x^2+(2x)/5-2x+1/3$

Simplify the terms with $x$ $x$ .

$0 = x^{2} + \frac{2 x}{5} - \frac{10 x}{5} + \frac{1}{3}$ $0=x^2+(2x)/5-(10x)/5+1/3$

$0 = x^{2} - \frac{8 x}{5} + \frac{1}{3}$ $0=x^2-(8x)/5+1/3$

We could apply the quadratic formula to this, but it would be simpler if we eliminated the fractions. To do this, multiply both sides by $15$ $15$ .

$0 = 15 x^{2} - 24 x + 5$ $0=15x^2-24x+5$

We can now use the quadratic formula, which states that for a quadratic function $0 = a x^{2} + b x + c$ $0=ax^2+bx+c$ , the roots are

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

Thus, $a = 15, b = - 24, c = 5$ $a=15,b=-24,c=5$ , so

$x = \frac{- (- 24) \pm \sqrt{{(- 24)}^{2} - (4 \times 15 \times 5)}}{2 \times 15}$ $x=(-(-24)+-sqrt((-24)^2-(4xx15xx5)))/(2xx15)$

$x = \frac{24 \pm \sqrt{576 - 300}}{30}$ $x=(24+-sqrt(576-300))/30$

$x = \frac{24 \pm \sqrt{276}}{30}$ $x=(24+-sqrt276)/30$

$x = \frac{24 \pm \sqrt{2^{2} \times 69}}{30}$ $x=(24+-sqrt(2^2xx69))/30$

$x = \frac{24 \pm 2 \sqrt{69}}{30}$ $x=(24+-2sqrt69)/30$

$x = \frac{12 \pm \sqrt{69}}{15}$ $x=(12+-sqrt69)/15$

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How to solve 2x-1/3 = x^2+2x/5 using the quadratic formula?

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