find x. there will be two values
2006-08-04 00:41:14 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(a^2+b^2)/2, (a^2-b^2)/2
2006-08-04 02:50:35 · answer #1 · answered by lose control 2 · 0⤊ 0⤋
You can find x only in terms of a and b. Using the quadratic formula
x=(-(4a) ± â((4a)^2 - 4*4*(a^4-b^4)))/2*4
2006-08-04 00:56:30 · answer #2 · answered by kangaruth 3 · 0⤊ 0⤋
4x^2 + (-4a^2)x + (a^4 - b^4)
x = (-b ± sqrt(b^2 - 4ac))/(2a)
a = 4
b = -4a^2
c = (a^4 - b^4)
x = (-(-4a^2) ± sqrt((-4a^2)^2 - 4(4)(a^4 - b^4)))/(2(4))
x = (4a^2 ± sqrt(16a^4 - 16a^4 + 4b^4))/8
x = (4a^2 ± sqrt(4b^4))/8
x = (4a^2 ± 2b^2)/8
x = 2(2a^2 ± b^2)/8
x = (1/4)(2a^2 ± b^2)
ANS :
x = (1/4)(2a^2 - b^2) or (1/4)(2a^2 + b^2)
2006-08-04 01:42:39 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
x = (4a^2 +/- sqrt( 16a^4 - 16(a^4 - b^4)) )/8
ax^2 + bx + c = 0
x = (-b +/- sqrt( b^2 -4ac) )/2a
2006-08-04 00:49:02 · answer #4 · answered by Auggie 3 · 0⤊ 0⤋
question is not defined enough, there are 3 unknowns
2006-08-04 00:45:02 · answer #5 · answered by bs 2 · 0⤊ 0⤋
you can't
2006-08-04 00:46:33 · answer #6 · answered by Roonal.18™ 3 · 0⤊ 0⤋