How do you solve $8 x^{2} + 4 = - 33 x$ $8x^2+4=-33x$ ?

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How do you solve $8 x^{2} + 4 = - 33 x$ $8x^2+4=-33x$ ?

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Answer 1 · 2016-04-05T17:53:52

The solutions are:
$x = - \frac{1}{8}$ $x = color(blue)(-1/8$

$x = - 4$ $x= color(blue)( - 4$

Explanation:

$8 x^{2} + 4 = - 33 x$ $8x^2 + 4 = -33x$

$8 x^{2} + 4 + 33 x = 0$ $8x^2 + 4 + 33x = 0$

$8 x^{2} + 33 x + 4 = 0$ $8x^2 + 33x + 4 = 0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:

$a = 8, b = 33, c = 4$ $a=8, b=33, c=4$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (33)^2-(4* 8 * 4)$

$= 1089 -128= 961$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$=(-33+-sqrt961)/(2*8)$

$=(-33+-31)/(16)$

$x =(-33+31)/(16) = -2/16 = color(blue)(-1/8$

$x =(-33-31)/(16) = -64/16 = color(blue)( - 4$

Answer 2 · 2016-04-06T16:02:16

1/8 and 4

Explanation:

$y = 8x^2 + 33x + 4 = 0$
Use the new Transforming Method (Google and Yahoo Search).
Transformed equation: $y' = x^2 + 33x + 32$ .
Since a - b + c = 0, use shortcut. Two real roots of y' are: -1 and (-c/a = -32).
Back to y, divide these real roots by (a = 8) to get the 2 real roots of y --> : $x1 = -1/8$ and $x2 = -32/8 = -4.$

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How do you solve $8 x^{2} + 4 = - 33 x$ $8x^2+4=-33x$ ?

2 Answers

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