4x^4-18x^2=0. I have to find out what x equals by factoring. What I did was
Factor 2x^2(2x^2+-9)=0 then 2x^2(2x+3)(x+-3), resulting in -1.5 and three. But these answers are wrong according to my book. Can you show me step by step how to do this correctly? It would be a huge help!
They are wrong.
2x^2(2x^2-9)=0
2x^2(xsqrt2-3)(xsqrt2+3)=0
so the roots are 0,0,3/sqrt2, -3/sqrt2)
Why did you square root the second line? Would you please be able to explain? Thanks!
Of course! Let's go through the factoring process step by step to find the correct values of x.
The equation you're given is:
4x^4 - 18x^2 = 0
First, notice that both terms have a common factor of 2x^2, so we can factor it out:
2x^2(2x^2 - 9) = 0
Now, we have two factors to consider: 2x^2 and (2x^2 - 9).
1. Setting the first factor equal to zero, we get:
2x^2 = 0
Dividing both sides by 2, we find:
x^2 = 0
Taking the square root of both sides, we have:
x = 0
So, one possible solution is x = 0.
2. For the second factor, (2x^2 - 9) = 0, we can try to solve it using the quadratic formula (ax^2 + bx + c = 0):
a = 2, b = 0, c = -9
The quadratic formula states that:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values, we have:
x = (0 ± √(0^2 - 4*2*(-9))) / (2*2)
Simplifying inside the square root:
x = (0 ± √(0 + 72)) / 4
x = (0 ± √(72)) / 4
x = (0 ± 6√2) / 4
Simplifying further:
x = ±3√2 / 2
This means we have two more solutions:
x = 3√2 / 2 and x = -3√2 / 2
So, the correct values of x for the equation 4x^4 - 18x^2 = 0 are x = 0, x = 3√2 / 2, and x = -3√2 / 2.
I hope this helps clarify the factoring process! Let me know if you have any further questions.