How do you solve $4 c^{2} + 10 c = - 7$ $4c^2 + 10c = -7$ by completing the square?

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Answer 1 · 2018-03-20T21:28:05

$c = - \frac{5}{4} \pm \frac{\sqrt{- 3}}{4}$ $c= -5/4 +- sqrt(-3)/4$

Factor out the coefficient of $x^{2}$ $x^2$ , then work with the quadratic expression which has a coefficient of $x^{2}$ $x^2$ equal to $1$ $1$
Check the coefficient of $x$ $x$ in the new quadratic expression and take half of it - this is the number that goes into the complete square bracket
Balance the constant term by subtracting the square of the number from step 2 and putting in the constant from the quadratic expression
The rest is arithmetic that may often involve fractions

Re-write the equations as

$4 c^{2} + 10 c + 7 = 0$ $4c^2+10c+7 = 0$

$4 (c^{2} + \frac{10 c}{4} + \frac{7}{4}) = 0$ $4(c^2 + (10c)/4+7/4) =0$

$4 [{(c + \frac{10}{4 \cdot 2})}^{2} - {(\frac{5}{4})}^{2} + \frac{7}{4}] = 0$ $4[(c+10/(4*2))^2 - (5/4)^2 + 7/4] =0$

$4 [{(c + \frac{5}{4})}^{2} - \frac{25}{16} + \frac{7}{4}] = 0$ $4[(c+5/4)^2 - 25/16 + 7/4] = 0$

$4 [{(c + \frac{5}{4})}^{2} + \frac{3}{16}] = 0$ $4[(c+5/4)^2 + 3/16] = 0$

$4 {(c + \frac{5}{4})}^{2} = - \frac{3}{16} \cdot 4$ $4(c+5/4)^2 = - 3/16 * 4$

$2 (c + \frac{5}{4}) = \pm \sqrt{- \frac{3}{4}}$ $2(c+5/4) = +- sqrt( - 3/4)$

$c = - \frac{5}{4} \pm \frac{\sqrt{- 3}}{4}$ $c= -5/4 +- sqrt(-3)/4$

This just presents the quadratic in an alternative format.

How do you solve 4c2+10c=−74c^2 + 10c = -7 by completing the square?