solve this equation by using the quadratic form. x^4-81=0?
回答 (11)
2009-03-02 1:34 pm
✔ 最佳答案
This ends up being the different of two squares (a few times over)So first:
x^4 - 81 = 0
Factors to:
(x² + 9)(x² - 9) = 0
And factors again, we have a second set of squares:
(x² + 9)(x + 3)(x - 3) = 0
Now that we're fully factored, if any one of them is equal to 0, the whole thing is equal to 0, so we solve each for 0:
x² + 9 = 0 and x + 3 = 0 and x - 3 = 0
x² = -9 and x = -3 and x = 3
x = ±√(-9) and x = ±3
x = ±3i and ±3
2016-05-24 2:02 pm
Quadratic formula: [-b ± √(b^2 - 4ac)]/2a To get the values for a, b, and c in the quadratic formula, the equations has to equal 0. 1.) 3m^2 + 6m = 3 (subtract 3 from both sides) 3m^2 + 6m - 3 = 0 (factor out 3) 3(m^2 + 2m - 1) = 0 ===> a = 1, b = 2, c = -1 *You could have also used 3 for a, 6 for b, and -3 for c instead of factoring out the 3 since the equation still equaled 0. However, I went ahead and factored out the 3 because it's just easier to work with smaller numbers. m = [-b ± √(b^2 - 4ac)]/2a m = [-2 ± √(2^2 - 4(1)(-1))]/2(1) m = [-2 ± √(4 + 4)]/2 m = [-2 ± √(8)]/2 m = [-2 ± √(4*2)]/2 m = [-2 ± 2√(2)]/2 m = -1 ± √(2) <===ANSWER 2.) x^2 + 5x - 4 = 0 ===> a = 1, b = 5, c = -4 x = [-b ± √(b^2 - 4ac)]/2a x = [-5 ± √(5^2 - 4(1)(-4))]/2(1) x = [-5 ± √(25 + 16)]/2 x = [-5 ± √(41)]/2 <===ANSWER
2009-03-02 2:31 pm
x^4 - 81 = 0
(x^2)^2 - 9^2 = 0
(x^2 + 9)(x^2 - 9) = 0
(x^2 + 9)(x^2 - 3^2) = 0
(x^2 + 9)(x + 3)(x - 3) = 0
x^2 + 9 = 0
x^2 = -9
x = ±√(-9)
x = ±√(i^2 * 3^2)
x = ±3i
x + 3 = 0
x = -3
x - 3 = 0
x = 3
∴ x = ±3i , ±3
(x^2)^2 - 9^2 = 0
(x^2 + 9)(x^2 - 9) = 0
(x^2 + 9)(x^2 - 3^2) = 0
(x^2 + 9)(x + 3)(x - 3) = 0
x^2 + 9 = 0
x^2 = -9
x = ±√(-9)
x = ±√(i^2 * 3^2)
x = ±3i
x + 3 = 0
x = -3
x - 3 = 0
x = 3
∴ x = ±3i , ±3
2009-03-02 1:39 pm
x^4-81=0
x^4 = 81
x = 4√81 = 3
x^4 = 81
x = 4√81 = 3
2009-03-02 1:37 pm
x^4=81
x^4=3^4
x=3
=)
x^4=3^4
x=3
=)
2009-03-02 1:37 pm
Hence, x ^ 4 = 81
i.e. (x ^ 2)^2 = SquareRoot(81)
i.e. (x ^ 2) = +/- 9
discaring negative value (-9) since it cannot be the value of squaring a number.
i.e. x ^ 2 = +9
i.e SquareRoot((x ^ 2)) = SquareRoot(9)
i.e. x= +/- 3
i.e. (x ^ 2)^2 = SquareRoot(81)
i.e. (x ^ 2) = +/- 9
discaring negative value (-9) since it cannot be the value of squaring a number.
i.e. x ^ 2 = +9
i.e SquareRoot((x ^ 2)) = SquareRoot(9)
i.e. x= +/- 3
2009-03-02 1:36 pm
This equation is the difference of 2 squares.
(x^2-9)(x^2+9)
Which is another didfference of squares.
(x-3)(x+3)(x^2+9)
Then set each part equal to 0.
x-3=0 x=3
x+3=0 x=-3
You cannot set the 3rd part equal to 0, because it wont work.
(x^2-9)(x^2+9)
Which is another didfference of squares.
(x-3)(x+3)(x^2+9)
Then set each part equal to 0.
x-3=0 x=3
x+3=0 x=-3
You cannot set the 3rd part equal to 0, because it wont work.
2009-03-02 1:35 pm
x^4-81=0
(x^2)^2-9^2=0 (x^2-y^2=(x-y)(x+y))
(x^2-9)(x^2+9)=0
x^2+9>=9
x^2-9=0
x=3 or -3
Maybe you donot the means of +-3i
(x^2)^2-9^2=0 (x^2-y^2=(x-y)(x+y))
(x^2-9)(x^2+9)=0
x^2+9>=9
x^2-9=0
x=3 or -3
Maybe you donot the means of +-3i
2009-03-02 1:35 pm
x^4-81=0
(x²+9)(x²-9) = 0
x = ±3, ±3i
.
(x²+9)(x²-9) = 0
x = ±3, ±3i
.
2009-03-02 1:34 pm
(x^2 + 3^2) (x^2 - 3^2)
(x^2 + 3^2) (x + 3) (x-3)
(x^2 + 3^2) (x + 3) (x-3)
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