How do you solve $8 x^{2} + 14 x + 5 = 0$ $8x^2 + 14x + 5 = 0$ by completing the square?

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How do you solve $8 x^{2} + 14 x + 5 = 0$ $8x^2 + 14x + 5 = 0$ by completing the square?

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Answer 1 · 2018-05-08T18:52:14

$x = - \frac{5}{4} or x = - \frac{1}{2}$ $x=-5/4" or "x=-1/2$

Explanation:

$using the method of completing the square$ $"using the method of "color(blue)"completing the square"$

$∙ the coefficient of the x^{2} term must be 1$ $• " the coefficient of the "x^2" term must be 1"$

$factor out 8$ $"factor out 8"$

$\Rightarrow 8 (x^{2} + \frac{7}{4} x + \frac{5}{8}) = 0$ $rArr8(x^2+7/4x+5/8)=0$

$∙ add/subtract {(\frac{1}{2} coefficient of the x-term)}^{2} to$ $• "add/subtract "(1/2"coefficient of the x-term")^2" to"$
$x^{2} + \frac{7}{4} x$ $x^2+7/4x$

$8 (x^{2} + 2 (\frac{7}{8}) x + \frac{49}{64} - \frac{49}{64} + \frac{5}{8}) = 0$ $8(x^2+2(7/8)xcolor(red)(+49/64)color(red)(-49/64)+5/8)=0$

$\Rightarrow 8 {(x + \frac{7}{8})}^{2} + 8 (- \frac{49}{64} + \frac{5}{8}) = 0$ $rArr8(x+7/8)^2+8(-49/64+5/8)=0$

$\Rightarrow 8 {(x + \frac{7}{8})}^{2} - \frac{9}{8} = 0$ $rArr8(x+7/8)^2-9/8=0$

$\Rightarrow 8 {(x + \frac{7}{8})}^{2} = \frac{9}{8}$ $rArr8(x+7/8)^2=9/8$

$divide both sides by 8$ $"divide both sides by 8"$

$\Rightarrow {(x + \frac{7}{8})}^{2} = \frac{9}{64}$ $rArr(x+7/8)^2=9/64$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x + \frac{7}{8})}^{2}} = \pm \sqrt{\frac{9}{64}} \leftarrow note plus or minus$ $sqrt((x+7/8)^2)=+-sqrt(9/64)larrcolor(blue)"note plus or minus"$

$\Rightarrow x + \frac{7}{8} = \pm \frac{3}{8}$ $rArrx+7/8=+-3/8$

$subtract \frac{7}{8} from both sides$ $"subtract "7/8" from both sides"$

$\Rightarrow x = - \frac{7}{8} \pm \frac{3}{8}$ $rArrx=-7/8+-3/8$

$\Rightarrow x = - \frac{7}{8} - \frac{3}{8} = - \frac{10}{8} = - \frac{5}{4}$ $rArrx=-7/8-3/8=-10/8=-5/4$

$or x = - \frac{7}{8} + \frac{3}{8} = - \frac{4}{8} = - \frac{1}{2}$ $"or "x=-7/8+3/8=-4/8=-1/2$

Answer 2 · 2018-05-08T18:58:25

$x = - \frac{10}{8} = - \frac{5}{4}$ $x= -10/8=-5/4$
$x = - \frac{4}{8} = - \frac{1}{2}$ $x=-4/8=-1/2$

Explanation:

Given: $y = 0 = 8 x^{2} + 14 x + 5$ $color(green)(y=0=8x^2+14x+5)$

Write as:
$0 = 8 (x^{2} + \frac{14}{8} x) + 5 ddd \to ddd 0 = 8 (x^{2} + \frac{7}{4} x) + 5$ $color(green)(0=8(x^2+14/8x)+5 color(white)("ddd")->color(white)("ddd")0=8(x^2+7/4x)+5)$

The 'perfect' square related to $x^{2} + \frac{7}{4} x is {(x + \frac{7}{8})}^{2}$ $x^2+7/4x " is " (x+7/8)^2$

but $8 {(x + \frac{7}{8})}^{2}$ $8(x+7/8)^2$ introduces the value of $8 \times {(\frac{7}{8})}^{2}$ $color(red)(8xx(7/8)^2)$ that is not in the original equation. We put it in so we have to take it out.

$0 = 8 (x^{2} + \frac{7}{4} x) + 5$ $color(green)(0=8(x^2+7/4x)+5 )$

$0 = 8 {(x + \frac{7}{8})}^{2} - [8 \times {(\frac{7}{8})}^{2}] + 5$ $color(green)(0= 8(x+7/8)^2color(red)(-[8xx(7/8)^2 ] )+5 )$

$0 = 8 {(x + \frac{7}{8})}^{2} - \frac{9}{8}$ $color(green)(0=8(x+7/8)^2-9/8 )$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Check$ $color(blue)("Check")$

$8 {(x + \frac{7}{8})}^{2} - \frac{9}{8}$ $8(x+7/8)^2-9/8$
$8 (x^{2} + \frac{7}{4} x + \frac{49}{64}) - \frac{9}{8}$ $8(x^2+7/4x +49/64)-9/8$

$8 x^{2} + 14 x + \frac{49}{8} - \frac{9}{8}$ $8x^2+14x+49/8-9/8$

$8 x^{2} + 14 x + \frac{40}{8}$ $8x^2+14x +40/8$

$8 x^{2} + 14 x + 5$ $8x^2+14x+5$ as required!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$0 = 8 {(x + \frac{7}{8})}^{2} - \frac{9}{8} dd \to dd + \frac{9}{8} = 8 {(x + \frac{7}{8})}^{2}$ $color(green)(0=8(x+7/8)^2-9/8color(white)("dd") ->color(white)("dd")+9/8=8(x+7/8)^2$

$ddddddddddddddddddd \to dddd \frac{9}{64} = {(x + \frac{7}{8})}^{2}$ $color(green)( color(white)("ddddddddddddddddddd")->color(white)("dddd") 9/64=(x+7/8)^2)$

$ddddddddddddddddddd \to d.d \pm \frac{3}{8} = d x + \frac{7}{8}$ $color(green)(color(white)("ddddddddddddddddddd")->color(white)("d.d")+-3/8=color(white)("d")x+7/8)$

$ddddddddddddddddddd \to ddd.dd x = - \frac{7}{8} \pm \frac{3}{8}$ $color(green)(color(white)("ddddddddddddddddddd")->color(white)("ddd.dd")x=-7/8+-3/8)$

$x = - \frac{10}{8} = - \frac{5}{4}$ $x= -10/8=-5/4$
$x = - \frac{4}{8} = - \frac{1}{2}$ $x=-4/8=-1/2$

Tony B

Answer 3 · 2018-05-08T19:01:33

$x = - \frac{1}{2} or x = - \frac{5}{4}$ $x=-1/2 or x=-5/4$

Explanation:

We have,

$8 x^{2} + 14 x + 5 = 0$ $8x^2+14x+5=0$

$\Rightarrow 8 x^{2} + 14 x = - 5$ $=>8x^2+14x=-5$

$=>8x^2+14x+K=K-5...to(A)$

Now we have to find $K$ ,such that, $(8x^2+14x+K)$ is a square.

Here,

$I^(st)term=8x^2$

$II^(nd) term=14x$

$III^(rd)term=K$

Formula for $color(red)(III^(rd)term=(II^(nd)term)^2/(4xxI^(st)term)...to(psi)$

$i.e. K=(14x)^2/(4xx8x^2)=(196x^2)/(32x^2)=49/8$

So,from $(A)$ ,we get

$8x^2+14x+49/8=49/8-5$

$(sqrt8x)^2+2(sqrt8x)(7/sqrt8)+(7/sqrt8)^2=9/8$

$(sqrt8x+7/sqrt8)^2=(3/sqrt8)^2$

$=>sqrt8x+7/sqrt8=+-3/sqrt8$

$=>8x+7=+-3$

$=>8x+7=3 or 8x+7=-3$

$=>8x=3-7 or 8x=-3-7$

$=>8x=-4 or 8x=-10$

$=>x=-1/2 or x=-5/4$

$=>x=-0.5 or x=-1.25$

Note: Formula $color(red)((psi)$ for $color(red)(III^(rd)term$ , is always true.

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