how would i solve this
x to the four - 15 x^ + 56 = 0
^ = squared
algebra equation, help??
2008-08-15 7:01 am
回答 (7)
2008-08-15 7:08 am
✔ 最佳答案
x^4 - 15x^2 + 56 = 0Look 56 = 7 x 8 & 15 = 7 + 8
so,
x^4 - (7 + 8)x^2 + 56 = 0
=> x^4 - 7x^2 - 8x^2 + 56 = 0
=> x^2(x^2 - 7) - 8 (x^2 - 7) = 0
=>(x^2 - 7)(x^2 - 8)= 0
=> x^2 - 7 = 0 or x^2 - 8 = 0
=> x = +/- sqrt(7) or x = +/- sqrt(8)
ANSWERS : x = +/- sqrt(7) , +/-sqrt(8)
2008-08-15 2:05 pm
(x^2-8)(x^2-7)=0
x^2-8=0, x^2-7=0
x^2=8, x^2=7
x=+-sqrt8, x=+-sqrt7
4 answers
make it a good day
x^2-8=0, x^2-7=0
x^2=8, x^2=7
x=+-sqrt8, x=+-sqrt7
4 answers
make it a good day
2008-08-15 7:25 pm
x^4 - 15x^2 + 56 = 0
x^4 - 7x^2 - 8x^2 + 56 = 0
(x^4 - 7x^2) - (8x^2 - 56) = 0
x^2(x^2 - 7) - 8(x^2 - 7) = 0
(x^2 - 7)(x^2 - 8) = 0
x^2 - 7 = 0
x^2 = 7
x = 屉7
x^2 - 8 = 0
x^2 = 8
x = 屉8
x = 2(屉2)
â´ x = ±â7 , 2(±â2)
x^4 - 7x^2 - 8x^2 + 56 = 0
(x^4 - 7x^2) - (8x^2 - 56) = 0
x^2(x^2 - 7) - 8(x^2 - 7) = 0
(x^2 - 7)(x^2 - 8) = 0
x^2 - 7 = 0
x^2 = 7
x = 屉7
x^2 - 8 = 0
x^2 = 8
x = 屉8
x = 2(屉2)
â´ x = ±â7 , 2(±â2)
2008-08-15 2:16 pm
Take X^2 =Y solve for Y then X=+ or - 8^1/2 and + or - 7^1/2
2008-08-15 2:10 pm
This is a standard form of a quadratic equation:
Ax(squared) + Bx + C = 0
It is solved by using the quadratic formula; see source below.
Ax(squared) + Bx + C = 0
It is solved by using the quadratic formula; see source below.
參考: Check out the site, "Answers.com," and type in "quadratic equation."
It shows you the steps to answer your question using the quadratic formula.
2008-08-15 2:08 pm
x^4 - 15 x ² + 56 = 0
y ² - 15 y + 56 = 0
(y - 8)(y - 7) = 0
y = 8 , y = 7
x ² = 8 , x ² = 7
x = ± 2â2 , x = ±â7
y ² - 15 y + 56 = 0
(y - 8)(y - 7) = 0
y = 8 , y = 7
x ² = 8 , x ² = 7
x = ± 2â2 , x = ±â7
2008-08-15 2:08 pm
(x^2-7)(x^2-8) = 0
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