Solve for x in equation
b²x²+b(a-c)x-ac=0
Give x in terms of a,b and c.
quadratic equation
2008-06-19 1:40 am
回答 (3)
2008-06-19 1:58 am
✔ 最佳答案
b²(a - c)² - 4(b²)(-ac)= b²a² - 2b²ac + b²c² + 4b²ac
=b²a² + 2b²ac + b²c²
= b²(a + c)²
x = {-b(a - c) + sqrt[b²(a - c)² - 4(b²)(-ac)]}/ (2b²)
or x = {-b(a - c) - sqrt[b²(a - c)² - 4(b²)(-ac)]}/ (2b²)
x = {-b(a - c) + b(a + c)}/ (2b²)
or x = {-b(a - c) - b(a + c)}/ (2b²)
x = 2bc/ (2b²) or x = -2ab/ (2b²)
x = c/b or x = - a / b
2008-06-19 2:05 am
By quadratic formula
x={-b(a-c)±sqrt([b(a-c)]²-4b²(-ac))}/2b²
={b(c-a)±sqrt[b²(a²-2ac+c²)+4acb²]}/2b²
={b(c-a)±sqrt[b²(a²+2ac+c²)]}/2b²
={b(c-a)±b(a+c)}/2b²
=[b(c-a)+b(a+c)]/2b² or [b(c-a)-b(a+c)]/2b²
=c/b or -a/b
x={-b(a-c)±sqrt([b(a-c)]²-4b²(-ac))}/2b²
={b(c-a)±sqrt[b²(a²-2ac+c²)+4acb²]}/2b²
={b(c-a)±sqrt[b²(a²+2ac+c²)]}/2b²
={b(c-a)±b(a+c)}/2b²
=[b(c-a)+b(a+c)]/2b² or [b(c-a)-b(a+c)]/2b²
=c/b or -a/b
2008-06-19 2:03 am
b²x² b(a-c)x-ac=0
x= { -b(a-c) /- root{[b(a-c)]² 4ac(b²)}} /2b²
= { b(c-a) /- root{ (ab)² - 2ac(b)² (bc)² 4ac(b)² } } /2b²
= { b(c-a) /- root { (ab bc)² } } /2b²
= [b(c-a) /- b(a c)] / 2b²
= c/b or -a/b
x= { -b(a-c) /- root{[b(a-c)]² 4ac(b²)}} /2b²
= { b(c-a) /- root{ (ab)² - 2ac(b)² (bc)² 4ac(b)² } } /2b²
= { b(c-a) /- root { (ab bc)² } } /2b²
= [b(c-a) /- b(a c)] / 2b²
= c/b or -a/b
收錄日期: 2021-04-27 15:01:37
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