How do you solve $- 3 x^{2} + 7 x = - 5$ $-3x^2+7x= -5$ by completing the square?

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How do you solve $- 3 x^{2} + 7 x = - 5$ $-3x^2+7x= -5$ by completing the square?

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Answer 1 · 2018-06-24T17:55:35

$x = \frac{7 \pm \sqrt{109}}{6}$ $x=(7+-sqrt109)/6$

Explanation:

You should always start by what you want to reach, namely an expression on the form of
${(x - a)}^{2} = b$ $(x-a)^2=b$
But first, to get some idea, let's start with a graph of the function
$f (x) = - 3 x^{2} + 7 x + 5$ $f(x)= -3x^2+7x+5$ :
enter image source here

We want ${(x - a)}^{2} = x^{2} - 2 a x + a^{2}$ $(x-a)^2=x^2-2ax+a^2$ on the left side
Our expression has
$- 3 x^{x} + 7 x = - 5$ $-3x^x+7x=-5$
Let's simplify by dividing the whole on $- 3$ $-3$ :
$x^{2} - \frac{7}{3} x = \frac{5}{3}$ $x^2-7/3x=5/3$

Therefore $- 2 a x = - \frac{7}{3} x$ $-2ax=-7/3x$ or $2 a = \frac{7}{3}$ $2a=7/3$ , $a = \frac{7}{6}$ $a=7/6$
Add $a^{2} = {(\frac{7}{6})}^{2}$ $a^2=(7/6)^2$ on both sides:
$x^{2} - \frac{7}{3} x + {(\frac{7}{6})}^{2} = \frac{5}{3} + \frac{49}{6^{2}}$ $x^2-7/3x+(7/6)^2=5/3+49/6^2$
${(x - \frac{7}{6})}^{2} = \frac{5 \cdot 12 + 49}{6^{2}} = \frac{109}{6^{2}}$ $(x-7/6)^2=(5*12+49)/6^2=109/6^2$
$x - \frac{7}{6} = \pm \frac{\sqrt{109}}{6}$ $x-7/6=+-sqrt109/6$
$x = \frac{7 \pm \sqrt{109}}{6}$ $x=(7+-sqrt109)/6$

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How do you solve $- 3 x^{2} + 7 x = - 5$ $-3x^2+7x= -5$ by completing the square?

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