What are the solutions of the equation x^4-5x-^2-14=0

1 answer:

8 0

So you want to start to factor so you get (x²+2)(x²-7) = 0. Now set each () to 0, so you get (x²+2) = 0 and (x²-7)=0. The first ends with +√2i and -√2i, the second ends with +√7 and -√7

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Which of the point-slope equations below are correct for the line that passes through points (-2, -2) and (2, 1)?

Ne4ueva [31]

Find slope first:
m =y2-y1/x2-x1
m = 1--2/2--2
m = 3/4

Select a point, insert your slope, & put it into point slope form:
y-y1 =m(x-x1)

Your final answers are:

y-1=3/4(x-2)
OR
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3 years ago

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aksik [14]

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find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. Can someone please show me ho

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For a quadratic of the form $f(x)=ax^2+bx+c$ , we have the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$ ,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given $2x^2-16x+27=0$ , we can get that $a=2, \ b=-16$ , and $c=27$ . We substitute these numbers into the quadratic formula above.

$x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)}$

$x=\dfrac{16 \pm \sqrt{(256 -216)} }{4}$

$x=\dfrac{16 \pm \sqrt{40} }{4}$

$x=\dfrac{16 \pm 2\sqrt{10} }{4}$

$x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}$

This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the $\pm$ symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.

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4 years ago