Solve 2x^4+16x^2=40?

Answer 1

Put x² = y

Then you'll have

2y² +16y = 40
2y² +16y - 40 = 0
2(y² +8y - 20) = 0
2(y + 10)(y - 2) = 0

Now it's time to substitute y = x².

2(x² + 10)(x² - 2) = 0

Factorize

2(x² + 10)(x - √2)(x + √2) = 0

The solutions are x = ±√2.

x² + 10 cannot be factorised in real numbers, but if you are allowed complex solutions, then the third and the fourth solutions are:

x = ± i √10

Hope this helps.

Answer 2

First, get everything to one side:
2x^4 + 16x - 40 = 0
Just to make things easier, divide by 2:
x^4 + 8x - 20 = 0
Now, think of two numbers whose sum is 8 and whose product is -20. I try to find the numbers that multiply to -20 first: -4, 5; -5, 4; 2, -10; -2, 10; 1, 20; -1, 20
Which pair of these add to 8? -2 and 10 do.
Now, factor out the polynomial:
(x^2 - 2)(x^2 + 10) = 0
Lastly, find when each of these factors equals zero. That is,
x^2 - 2 = 0 and x^2 + 10 = 0
x^2 = 2 x^2 = -10
For the first one, the solutions are +/- square root of 2.
For the second one, solutionas are +/- root of 10 times i, where i is the root of negative one.

Answer 3

Solve 2x^4+16x^2=40

2x^4+16x^2-40 = 0

Divide through by 2

x^4+8x^2-20 = 0

factor

(x^2+10)(x^2-2) = 0

x= + i(10)^0.5, - i(10)^0.5, +(2)^0.5, -(2)^0.5
.

Answer 4

assume x^2=y then
2y^2+16y=40 for x^4=y^2
2y^2+16y-40=0
divide by 2 then
y^2+8y-20=0
(y+10)(y-2)=0
y= -10 or y=2
for x^2 =y
x^2= -10 or x^2=2
we omit x^2= -10 no negative number's square root is a real number.
so x^2=2
x= + - squareroot 2

Answer 5

2(x^4 + 8x^2 - 20)
2(x^2 + 10)(x^2 - 2)

now solve the brackets

x^2 + 10 = 0
x^2 = -10
x = ±(-10)^1/2 = ±(10^1/2)i, where i is the imaginary number (-1)^1/2

second set
x^2 - 2 = 0
x^2 = 2
x = ±(2^1/2)

and as for the above answer, you can't just simply use y = x^2 since you will be still have to convert back from the parametrization you have done to get the roots for the original equation.

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nice edit after the fact for the above answer ;)