Math problem help!!!?

Question

can you help me solve this with whatever method?

4x^2-9=0

i think one of the solutions might be 3/4ths, but i'm not sure thanks.

Answer 1

Ok so because this 4x^2-9 is factorable, you can factor it into (2x+3)(2x-3).

Set each equation to zero.

2x+3=0 and 2x-3=0

2x=-3 and 2x=3

x=3/2 and -3/2 are your answers!

Answer 2

(2x -3) (2x+3) FOIL gives you 4x^2... the -3x and 3x cancel.. and the last gives you -9... answers are x=3/2 and -3/2

Answer 3

4x² - 9 = 0

4x² = 9

√(4x²) = √9

2x = 3

x = +/- 3/2

Answer: x = +/- 3/2 or +/- 1 1/2

Proof:

4([- 3/2]²) = 9

4(9/4) = 9

9 = 9

Answer 4

4 = 2^2

9 = 3^2

So you can write this as...

(2x)^2 - 3^2

This is the form of something called "The Difference of Squares".  You should learn to recoginze this formm as it is easily factored...

a^2 - b^2 = (a-b)(a+b)

so

(2x)^2 - 3^2 = (2x - 3)(2x + 3)

Your equation now reads...

(2x - 3)(2x + 3) = 0

By the zero property...

2x - 3 = 0 or 2x + 3 = 0

So

x = 3/2 or x = -3/2

Plug these values into the equation to check....

  

Answer 5

4x^2 -- 9 = 0 Or x^2 = 9/4 Or (x)^2 = (3/2)^2 whence x = 3/2, -- 3/2