find all the real zeros of the equation:2x^4-2x^2-40?

Question

please explain.......

Answer 1

Take out a 2: 2(x^4-x^2-20)
Factor: 2(x^2-5)(x^2+4)
Set each part equal to zero and solve (the parts are in the parenthesis)
x^2-5=0
x^2=5 (added 5 to both sides)
x=+or - the sqaure root of 5 (square rooted both sides.)
x^2+4=0
x^2=-4 (subtracted 4 from both sides)
x=+ or - 2i (square rooted both sides)

So the zeros are +or - the square root of 5. +or - 2i would not be a real zero.

Answer 2

2x^4 - 2x^2 - 40= 0
2 (x^4 - x^2 - 20) = 0
(x^4 - x^2 - 20) = 0
(x^2 - 5) (x^2 + 4) = 0

x^2 - 5 = 0 or x^2 + 4 = 0

x^2 = 5 or x^2 = -4

The real roots are:
x = the square root of 5 and
x = negative square root of 5.

[Imaginary roots are x = +/- 2i. In this case, since you're looking only for real roots, recognize that two of the roots are not real roots, and disregard these non-real roots. Since you have a fourth degree equation, you should have four zeros, although they will not be distinct (different) in every case.]

Answer 3

Let x^2 = y

2y^2 - 2y - 40
= 2y^2 - 10y + 8y - 40
= 2y(y - 5) + 8(y - 5)
= (2y + 8)(y - 5)
= 2(y + 4)(y - 5)

Set the expression to zero

2(y + 4)(y - 5) = 0
---> (y + 4)(y - 5) = 0

Either,
y + 4 = 0 (or) y - 5 = 0
y = -4 (or) y = 5

x^2 cannot be -4
Reason:
If x^2 = -4
x = sqrt (-4)
which is imaginary

So, x^2 = 5
---> x = +/- sqrt 5

(sqrt 5) and -(sqrt 5) are the zeros of the given expression (not equation!)

Answer 4

2x^4-2x^2-40=0
2(x^4-x^2-20)=0

-b +- sqrt(b^2 - 4ac) / 2a

Exponents are even so its okay to use it

1 +- sqrt( -1^2 - 4(1)(-20)) / 2

1 +- sqrt( 1 + 80) / 2
1 +- 9 / 2

(x^2 + 5)(x^2 - 4)
(x^2 + 5) has no real roots

(x^2 - 4) = (x+2)(x-2)

roots are x = -2,2