jk11 wrote:
Please tell me what is wrong with the following procedure:
Q. Find solutions of the following expression.
(¼)x^4 - (¾)x^2 - 1 = 0
=> (¼)x^4 - (¾)x^2 = 1 -----------------moving 1 to RHS
=> (¼)*(x^2) * (x^2 - 3) = 1 ---------- taking (¼)*(x^2) common
=> (x^2)*(x^2 -3) = 4 ---------------- mulitplying 4 on both sides
=> x^2 = 4 OR x^2 - 3 = 4
=> x= +/-2 OR x = +/-7
Thus, there are four solutions: +2, -2, +7, -7.
However, the correct answer is that there are only two solutions: +2,-2.
Hey
jk11you could have either substituted x^2 as some variable lets
say a, or you could have just used it as it is, said that, i took the
latter x^2 will be +ive, x^4 will be positive
x^4 - 3x^2 - 4 = 0
x^2 (x^2 -4) + 1 (x^2-4) = 0
two solutions x^2 = -1 or
x^2 = 4only
this is possible
x^2 = 4
two solutions -2 or 2
_________________
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Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.