Given the qintic equation below, solve it to find the values of x. 2x^5_6x^3_4x^2_2x+4 =0​

9514 1404 393

  x = {-0.5-√1.25, -0.5+√1.25, 2, -0.5+i√0.75, -0.5-i√0.75}

Step-by-step explanation:

I like to use a graphing calculator to find clues as to the roots of higher-degree polynomials. Here, we see that x=2 is the only real rational root. Dividing that out by synthetic division, we see the remaining quartic factor is ...

  2x^5 -6x^3 -4x^2 -2x +4 = 0

  2(x -2)(x^4 +2x^3 +x^2 -1) = 0

We can recognize that the quartic factor is actually the difference of two squares:

  x^4 +2x^3 +x^2 -1 = (x^2 +x)^2 -1 = 0

So it resolves to two quadratic factors.

  (x^2 +x +1)(x^2 +x -1) = 0

One will have real roots, as shown by the graph. The other will have complex roots.

  x^2 +x + 1/4 = 1 +1/4 . . . . complete the square for the factor with real roots

  (x +1/2)^2 = 5/4

  x = -1/2 ± √(5/4) . . . . . . irrational real roots

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  x^2 +x = -1 . . . . . . . . . . the quadratic factor with complex roots

  (x +1/2)^2 = -1 +1/4 . . . complete the square

  x = -1/2 ± i√(3/4) . . . . irrational complex roots

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In summary, the values of x that satisfy the equation are ...

 x = 2

  x = -1/2 ± √(5/4)

  x = -1/2 ± i√(3/4)

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

5+2x=a

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