更新1:
Thanks all!!!
Is it possible to solve this equation?
回答 (9)
2010-03-12 4:13 am
✔ 最佳答案
for real numbers no. but if you allow imaginary or complex numbers like i, then the answer is +-3i
2010-03-12 12:15 pm
x = -3i
x = 3i
good luck
x = 3i
good luck
2010-03-12 12:18 pm
no you can't. You can;t find the square root of a negative number. Try it on your calculator it will be wrote : ERROR.
Normally, you would have to put the 9 on the other side of the equal then it will be: x2 = 0-9 ,
so x2= -9 but you can't do the square root at this point, if you want this problem to work you will have to have a negative 9 in the first equation.
Normally, you would have to put the 9 on the other side of the equal then it will be: x2 = 0-9 ,
so x2= -9 but you can't do the square root at this point, if you want this problem to work you will have to have a negative 9 in the first equation.
2010-03-12 12:14 pm
there are no real solutions. If you subtract 9 from both sides, you'll get x^2 = -9. Take the square root of both sides, and you'll have x = sq.rt.(-9). You cannot take the square root of a negative number(unless you're in a higher level math class like pre-calc)
參考: I'm a high school math teacher
2010-03-12 12:14 pm
No
if x^2= -9 there is no number that will be negative when you square it
if x^2= -9 there is no number that will be negative when you square it
2010-03-12 12:14 pm
no its not
2010-03-12 12:31 pm
yes its -3 you subtract 9 from both sides 9-9 cancels out and 0-9= -9 then the opposite of ^2 is this little symbol â wich is -3 so x= -3
-3^2+9=0
hope this helps
-3^2+9=0
hope this helps
參考: taking algerbra in school now
2010-03-12 12:22 pm
Use the quadratic formula.
negative b plus or minus the sq. root of b^2 - 4ac all over 2a
in this case a=1 b=0 and c=9
x=-(0)+/- â[0^2-4(1)(9)]/(2*1)
x=[0+/-â(-36)]/2
x=(0+/-6i)/2
x= 3i or x= -3i
2 imaginary roots even though there are no real roots there is a way to solve this equation
negative b plus or minus the sq. root of b^2 - 4ac all over 2a
in this case a=1 b=0 and c=9
x=-(0)+/- â[0^2-4(1)(9)]/(2*1)
x=[0+/-â(-36)]/2
x=(0+/-6i)/2
x= 3i or x= -3i
2 imaginary roots even though there are no real roots there is a way to solve this equation
參考: Math tutor Calculus and Down
2010-03-12 12:21 pm
Well.. they're always possible, you're just going to have imaginary numbers:
x^2=-9
x=+/-3i
have a nice day :)
x^2=-9
x=+/-3i
have a nice day :)
參考: Grade 10 Math
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