How do you solve the quadratic equation by completing the square: $2 x^{2} - 7 x = 2$ $2x^2 - 7x = 2$ ?

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How do you solve the quadratic equation by completing the square: $2 x^{2} - 7 x = 2$ $2x^2 - 7x = 2$ ?

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Answer 1 · 2015-07-14T16:14:48

The solution is $x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$ $x = 7/4 +- (sqrt65)/4$ .

Explanation:

$2 (x^{2} - \frac{7}{2} x) = 2$ $2(x^2 - 7/2x) = 2$
$x^{2} - \frac{7}{2} x + \frac{49}{16} = 1 + \frac{49}{16}$ $x^2 - 7/2x + 49/16 = 1 + 49/16$
${(x - \frac{7}{4})}^{2} = \frac{65}{16}$ $(x - 7/4)^2 = 65/16$
$(x - \frac{7}{4}) = \pm \frac{\sqrt{65}}{4}$ $(x - 7/4) = +- (sqrt65)/4$

$x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$ $x = 7/4 +- (sqrt65)/4$

Answer 2 · 2015-07-15T02:03:37

$x = \frac{7 + \sqrt{65}}{4},$ $x=(7+sqrt65)/4,$ $\frac{7 - \sqrt{65}}{4}$ $(7-sqrt65)/4$

Explanation:

$2 x^{2} - 7 x = 2$ $2x^2-7x=2$

Divide both sides by $2$ $2$ .

$x^{2} - \frac{7}{2} x = 1$ $x^2-7/2x=1$

To complete the square means to force a perfect square trinomial on the left side of the equation in the form $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$ .

Divide the coefficient of the $x$ $x$ term by $2$ $2$ , square the result, and add to both sides of the equation.

$\frac{- 7}{2} \div 2 = (- \frac{7}{2}) \cdot \frac{1}{2} = - \frac{7}{4}$ $(-7)/2-:2=(-7/2)*1/2=-7/4$
${(- \frac{7}{4})}^{2} = \frac{49}{16}$ $(-7/4)^2=49/16$

$x^{2} - \frac{7}{2} x + \frac{49}{16} = 1 + \frac{49}{16}$ $x^2-7/2x+49/16=1+49/16$

The common denominator for $1$ $1$ and $\frac{49}{16}$ $49/16$ is $16$ $16$ . Multiply $1$ $1$ times $\frac{16}{16}$ $16/16$ , then add the two fractions.

$x^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{16}{16} + \frac{49}{16}$ $x^2-7/2x+49/16=16/16+49/16$ =

$x^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{65}{16}$ $x^2-7/2x+49/16=65/16$

We now have a perfect square trinomial on the left side, where $a = x$ $a=x$ and $b = \frac{7}{4}$ $b=7/4$ .

${(x - \frac{7}{4})}^{2} = \frac{65}{15}$ $(x-7/4)^2=65/15$

Take the square root of both sides.

$x - \frac{7}{4} = \pm \sqrt{\frac{65}{16}}$ $x-7/4=+-sqrt(65/16)$ =

$x - \frac{7}{4} = \pm \frac{\sqrt{65}}{4}$ $x-7/4=+-sqrt65/4$

Solve for $x$ $x$ .

$x = \frac{7}{4} \pm \frac{\sqrt{65}}{4}$ $x=7/4+-sqrt65/4$ =

$x = \frac{7 \pm \sqrt{65}}{4}$ $x=(7+-sqrt65)/4$

$x = \frac{7 + \sqrt{65}}{4}$ $x=(7+sqrt65)/4$ =

$x = \frac{7 - \sqrt{65}}{4}$ $x=(7-sqrt65)/4$

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2 Answers

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