Use the quadratic formula to solve:
x(x+5)=-9
ALGEBRA. Can anyone help?
2008-12-31 3:20 am
回答 (7)
2008-12-31 3:32 am
✔ 最佳答案
x(x+5)=-9x^2+5x=-9
x^+5x+9=0
a=1 b=5 c=9
x=-b+/-sqrt(b^2-4ac)/2a
=-5+/-sqrt[(5)^2-4(1)(9)]/2(1)
=-5+/-sqrt(25-36)/2
=-5+/-sqrt(-11)/2
=-5+/-isqrt(11)/2
x=-5+isqrt(11)/2 or x=-5-isqrt(11)/2 answer @(^_^)@
2008-12-31 4:49 pm
x(x + 5) = -9
x^2 + 5x + 9 = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 1
b = 5
c = 9
x = [-5 屉(25 - 36)]/2
x = [-5 屉(-11)]/2
x = [-5 ±iâ11]/2
â´ x = [-5 ±iâ11]/2
x^2 + 5x + 9 = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 1
b = 5
c = 9
x = [-5 屉(25 - 36)]/2
x = [-5 屉(-11)]/2
x = [-5 ±iâ11]/2
â´ x = [-5 ±iâ11]/2
2008-12-31 11:29 am
A=1 B=5 C=9
-5(plus or minus)(square root of) -11
all over 2(a)
and if you have gotten to imaginary #s yet,
-5(plus or minus) (i times the square root of)11
over 2
-5(plus or minus)(square root of) -11
all over 2(a)
and if you have gotten to imaginary #s yet,
-5(plus or minus) (i times the square root of)11
over 2
參考: Aced algebra in HS
2008-12-31 11:29 am
x^2+5x+9=0
a= x^2 b= 5x c= 9
+/-b sqrt b - 4(a)(c)/ 2a
plug in the following and you will get the answer.
a= x^2 b= 5x c= 9
+/-b sqrt b - 4(a)(c)/ 2a
plug in the following and you will get the answer.
2008-12-31 11:26 am
x(x + 5) = -9 First distribute
x^2 + 5x = -9 Put in standard form
x^2 + 5x + 9 = 0
a = 1
b = 5
c = 9
formula is the opposite of b plus minus the square root of b squared - 4 times a times c all divided by 2 times a
-5 +- sq rt of 5^2 - 4(1)(9) / 2(1)
-5 +- sq rt of 25 - 36 / 2
And I see right away that 25 - 36 is a negative number and it can't be square rooted so there is NO Solution!
x^2 + 5x = -9 Put in standard form
x^2 + 5x + 9 = 0
a = 1
b = 5
c = 9
formula is the opposite of b plus minus the square root of b squared - 4 times a times c all divided by 2 times a
-5 +- sq rt of 5^2 - 4(1)(9) / 2(1)
-5 +- sq rt of 25 - 36 / 2
And I see right away that 25 - 36 is a negative number and it can't be square rooted so there is NO Solution!
2008-12-31 11:24 am
x^2 + 5x + 9 = 0
x = [-5 +/- sqrt(5^2 - 4*9)]/2
......[-5 +/- sqrt(25 - 36)]/2
......[-5 sqrt(-11)]/2
......[-5 +/- i sqrt11]/2
x = [-5 +/- sqrt(5^2 - 4*9)]/2
......[-5 +/- sqrt(25 - 36)]/2
......[-5 sqrt(-11)]/2
......[-5 +/- i sqrt11]/2
2008-12-31 11:22 am
x^2+5x+9=0
Cannot be solved as
square root [b^2-4ac] = sq.root[-11]
-11 dont' have a square root
Cannot be solved as
square root [b^2-4ac] = sq.root[-11]
-11 dont' have a square root
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