Question

Solve for the roots in the equation below.
x^4 + 3x^2 - 4 = 0

Answer 1

take \(x^2 =t \)
then \(x^4 = (x^2)^2 = t^2\)

so you will get a quadratic in 't'
can you solve quadratics ?

Answer 2

no. is it 9 and 16

Answer 3

t^2 + 3t - 4 = 0
can you factor this ?

Answer 4

you will need to find 2 numbers with sum = +3 and product = -4

Answer 5

1 + 2.
3 + -7=-4

Answer 6

same 2 numbers should give sum = 3 and product of same 2 numbers should be -4
try to find such 2 numbers

Answer 7

I can't figure it out

Answer 8

the 2 numbers are 4 and -1 because

4 * -1 = -4
and
-1 + 4 = +3

Answer 9

Thank you. It's 4 x -1= -4 -1 + 4= 3

Answer 10

Is this the answer

Answer 11

no these 2 numbers go into 2 brackets as follows

(t + 4)(t - 1) = 0

Answer 12

now either t +4 = 0 or t - 1 = 0

so there are 2 values of t -4 and 1

Answer 13

but going back to the original question t = x^2

so x^2 = 1 or -4