2007-10-06 03:09:40 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x^2-9)(x^2-4)=0
(x-3)(x+3)(x+2)(x-2)=0
(x-3)=0 ; (x+3)=0; (x+2)=0; (x-2)=0
x = 3 ; x = -3; x= -2; x=2
2007-10-06 03:19:20 · answer #1 · answered by tootoot 3 · 0⤊ 0⤋
Let x^2 = y
The equation becomes:
y^2 - 13y + 36 = 0
Now that we have a basic quadratic equation, we can dispense with the formalities:
y^2 - 13y + 36 = 0
y^2 - 4y - 9y + 36 = 0 (Splitting the middle term)
(y - 4)(y - 9) = 0 (Factorizing)
y = 4, 9
x^2 = 4 (or) x^2 = 9
x = +/-2 (or) x = +-3
x = 2, -2, 3, -3
2, -2, 3, -3 are the roots of the given equation
2007-10-06 10:37:27 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 1⤋
X^4-13x^2+36 =0
remark ( x^2-6)^2 = x^4-12x^2+36
so x^4-13x^2+36 = (x^2-6)^2-x^2 we use (a-b)(a+b) =a^2-b^2
(x^2-x-6)(x^2+x-6)=x^4-13x^2+36
roots x(1) = (1-(1+6*4)^0.5)/2= -2
and x(2) =(1+(1+6*4)^0.5)/2= 3
the roots are -2 and +3
2007-10-06 10:21:42 · answer #3 · answered by maussy 7 · 0⤊ 1⤋
(x² - 9)(x² - 4) = 0
x = ± 3 , x = ± 2
2007-10-06 15:48:10 · answer #4 · answered by Como 7 · 1⤊ 1⤋
Its already solved...there's nothing more you can do.
2007-10-06 10:12:45 · answer #5 · answered by blue_eyed_blonde0611 3 · 0⤊ 1⤋