How do you solve $4 x^{2} - 4 x = 15$ $4x^2-4x=15$ ?

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How do you solve $4 x^{2} - 4 x = 15$ $4x^2-4x=15$ ?

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Answer 1 · 2018-06-04T17:20:33

$x_{1} = \frac{5}{2}$ $x_1=5/2$ or $x_{2} = - \frac{3}{2}$ $x_2=-3/2$

Explanation:

Writing your equation in the form
$4 x^{2} - 4 x - 15 = 0$ $4x^2-4x-15=0$
dividing by $4$ $4$
$x^{2} - x - \frac{15}{4} = 0$ $x^2-x-15/4=0$
using the quadratic formula
$x_{1, 2} = \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{15}{4}}$ $x_{1,2}=1/2pmsqrt(1/4+15/4)$
so we get

$x_{1} = \frac{5}{2}$ $x_1=5/2$

$x_{2} = - \frac{3}{2}$ $x_2=-3/2$

Answer 2 · 2018-06-04T18:36:06

$x = - \frac{3}{2} or x = \frac{5}{2}$ $x=-3/2" or "x=5/2$

Explanation:

$rearrange in standard form; a x^{2} + b x + c = 0$ $"rearrange in standard form ";ax^2+bx+c=0$

$subtract 15 from both sides$ $"subtract 15 from both sides"$

$4 x^{2} - 4 x - 15 = 0$ $4x^2-4x-15=0$

$using the a-c method to factor the quadratic$ $"using the a-c method to factor the quadratic"$

$the factors of the product 4 \times - 15 = - 60$ $"the factors of the product "4xx-15=-60$

$which sum to - 4 are + 6 and - 10$ $"which sum to - 4 are + 6 and - 10"$

$split the middle term using these factors$ $"split the middle term using these factors"$

$4 x^{2} + 6 x - 10 x - 15 = 0 \leftarrow factor by grouping$ $4x^2+6x-10x-15=0larrcolor(blue)"factor by grouping"$

$2 x (2 x + 3) - 5 (2 x + 3) = 0$ $color(red)(2x)(2x+3)color(red)(-5)(2x+3)=0$

$take out the common factor (2 x + 3)$ $"take out the "color(blue)"common factor "(2x+3)$

$(2 x + 3) (2 x - 5) = 0$ $(2x+3)(color(red)(2x-5))=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$2 x + 3 = 0 \Rightarrow x = - \frac{3}{2}$ $2x+3=0rArrx=-3/2$

$2 x - 5 = 0 \Rightarrow x = \frac{5}{2}$ $2x-5=0rArrx=5/2$

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How do you solve $4 x^{2} - 4 x = 15$ $4x^2-4x=15$ ?

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