How do you use the quadratic formula to solve the equation 3x^2=3x.?
回答 (15)
2009-07-16 5:42 pm
✔ 最佳答案
Hi,For this question, notice that 3 is a common factor on both sides of the equation and can be canceled to get:
x^2 = x
Now, subtract 'x' from both sides to get:
x^2 - x = 0
Notice that 'x' can be factored out of this equation to get:
x(x-1) = 0
When this occurs, we know that each factor must be set equal to zero to get:
x = 0
AND
x - 1 = 0 or x = 1.
I hope that helps you out! Please let me know if you have any other questions!
參考: College Calculus Student ; Math Tutor
2009-07-15 2:12 pm
First, to make your life easier, I would divide both sides by 3; the only point of this is to make the numbers smaller: x^2 = x
Your equation must be in the form ax^2 + bx + c = 0 in order to use the quadratic formula, so subtract x from both sides: x^2 – x = 0
Now you can use the quadratic formula: x = [–b ± √(b^2 – 4ac)]/2a
First find a, b, and c: a = 1; b = –1; c = 0
Now plug them into the formula: x = [1 ± √(1 – 0)]/2 = (1 ± 1)/2 = 2/2, 0/2 = 0, 1
Therefore, your answer is: x = 0, 1
Your equation must be in the form ax^2 + bx + c = 0 in order to use the quadratic formula, so subtract x from both sides: x^2 – x = 0
Now you can use the quadratic formula: x = [–b ± √(b^2 – 4ac)]/2a
First find a, b, and c: a = 1; b = –1; c = 0
Now plug them into the formula: x = [1 ± √(1 – 0)]/2 = (1 ± 1)/2 = 2/2, 0/2 = 0, 1
Therefore, your answer is: x = 0, 1
2009-07-15 6:27 pm
3 x^2 - 3 x = 0
x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ 3 ± √ (9) ] / 6
x = [ 3 ± 3 ] / 6
x = 1 , x = 0
CHECK
3 x ( x - 1 ) = 0
x = 0 , x = 1
x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ 3 ± √ (9) ] / 6
x = [ 3 ± 3 ] / 6
x = 1 , x = 0
CHECK
3 x ( x - 1 ) = 0
x = 0 , x = 1
2009-07-15 9:12 am
3x^2 = 3x
3x^2 - 3x + 0 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)
a = 3
b = -3
c = 0
x = [3 ±√(9 - 0)]/6
x = [3 ±√9]/6
x = [3 ±3]/6
x = [3 + 3]/6
x = 6/6
x = 1
x = [3 - 3]/6
x = 0
∴ x = 0, 1
3x^2 - 3x + 0 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)
a = 3
b = -3
c = 0
x = [3 ±√(9 - 0)]/6
x = [3 ±√9]/6
x = [3 ±3]/6
x = [3 + 3]/6
x = 6/6
x = 1
x = [3 - 3]/6
x = 0
∴ x = 0, 1
2009-07-15 8:00 am
To use the quadratic formula, the equation has to be in the form of:
Ax² + Bx + C = 0
So subtract 3x from each side to get it into this form:
3x² - 3x + 0 = 0
The quadratic formula states that:
x = (-B ± √(B² - 4AC) )/ 2A
So here, A = 3, B = -3, C = 0
Plug these into the quadratic formula:
x = (-(-3) ± √((-3)² - 4(3)(0) ))/ 2(3)
x = (3 ± √(9)) /6
x = (3 ± 3) /6
x = 0 or x = 1
Hope this helps!
Ax² + Bx + C = 0
So subtract 3x from each side to get it into this form:
3x² - 3x + 0 = 0
The quadratic formula states that:
x = (-B ± √(B² - 4AC) )/ 2A
So here, A = 3, B = -3, C = 0
Plug these into the quadratic formula:
x = (-(-3) ± √((-3)² - 4(3)(0) ))/ 2(3)
x = (3 ± √(9)) /6
x = (3 ± 3) /6
x = 0 or x = 1
Hope this helps!
2009-07-15 7:56 am
Start by making one side equal to 0 by subtracting 3x from both sides:
3x^2 - 3x = 0
By the quadratic formula, x = [-b +/-√(b^2 - 4ac)]/2a
where ax^2 + bx + c = 0
Here we have
3*x^2 + -3*x +0 = 0
so a = 3, b = -3 and c = 0
so x = [-(-3) +/- √((-3)^2 - 4 * 3 * 0)]/(2*3)
= (3+/-3)/6
= 0 or 1
3x^2 - 3x = 0
By the quadratic formula, x = [-b +/-√(b^2 - 4ac)]/2a
where ax^2 + bx + c = 0
Here we have
3*x^2 + -3*x +0 = 0
so a = 3, b = -3 and c = 0
so x = [-(-3) +/- √((-3)^2 - 4 * 3 * 0)]/(2*3)
= (3+/-3)/6
= 0 or 1
2009-07-16 8:38 pm
3x^2 = 3x
x = 0
x = 1
x = 0
x = 1
2009-07-15 3:06 pm
You would not use the quadratic formula to solve that equation. You would only use the quadratic formula if it cannot be factored.
3x² + 3x
3x² - 3x = 0
3x(x - 1) = 0
3x = 0 OR x - 1 = 0
x = 0 OR x = 1
3x² + 3x
3x² - 3x = 0
3x(x - 1) = 0
3x = 0 OR x - 1 = 0
x = 0 OR x = 1
2009-07-15 9:37 am
x=0 & x=1
2009-07-19 1:48 am
This is an easy one to solve without the formula, but you teacher probably wants to make sure you understand the concept of the formula, with a slightly trick question. Two steps you need to take away from this.
1. Get everything on one side so that it equals 0. So, subtract 3x from both sides to get 3x^2 - 3x = 0
2. You have a the 3 in 3x^2 and b, the -3 in -3x, but you don't see the third term for c. So, that means c = 0. Just like if you had 3x^2 - 3 = 0
then a would be 3, c could be -3 and b would be 0.
Hope that helps.
1. Get everything on one side so that it equals 0. So, subtract 3x from both sides to get 3x^2 - 3x = 0
2. You have a the 3 in 3x^2 and b, the -3 in -3x, but you don't see the third term for c. So, that means c = 0. Just like if you had 3x^2 - 3 = 0
then a would be 3, c could be -3 and b would be 0.
Hope that helps.
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