Quad. Equation isn't even playing by the rules?

Question

If we had.
3x²-5x+2=0
The rules says that if we do one thing to one side we must to the same thing to the other.
If we would get 3x² to just x², we would have to divide both sides by "3" "Once to one side than the other side" Like a regular equation.
ex.

3x²/3-5x=-2/3
would be.
x²-5x=-2/3
But to really solve it we must divide "5" by "3" too, it makes no sense.

How is that we must take -5 and divide it by 3???
Why isn't it playing by the rules?

Answer 1

Hi,

3x² - 5x + 2 = 0 Move constant to the other side.
3x² - 5x = -2

Divide every term by 3 so the x² term has a coefficient of 1.
Doing the SAME thing to EVERY term IS playing by the rules!

3x² - 5x = -2
3/3x² - 5/3x = -2/3
x² - 5/3x = -2/3

Multiply ½ times -5/3. Square the result and add it to both sides.
½ × -5/3 = -5/6. (-5/6)² = 25/36 Add this to both sides.

x² - 5/3x + 25/36 = -2/3 + 25/36

Factor the left side.and combine like terms on the right.

(x - 5/6)² = 1/36

Take the square root of both sides. Put ± on the right.

x - 5/6 = ± 1/6

Add 5/6 to both sides.

x = 5/6 ± 1/6

Solve for both answers.
x = 5/6 + 1/6 = 6/6 = 1 <== first answer
x = 5/6 - 1/6 = 4/6 = 2/3 <== second answer


I hope this helps!! :-)

Answer 2

Tiger: you are in the right track. The equation given,

3x^2 - 5x + 2 = 0

The only factor you must convert to one is the 3x^2...so, you divide by 3.

x^2 - 5/3x + 2/3 = 0

recall the form of the equation is

x^2 + 2ax + a^2 = 0

[x^2 - 2(5/6)x + (5/6)^2] + (2/3 -(5/6)^2) = 0

(x-5/6)^2 + 2/3 - 25/36 = 0

(x-5/6)^2 + [24/36 - 25/36]= 0

or (x-5/6)^2 = 1/36

Taking the square root on both sides we get

(x-5/6) = -1/6, 1/6

x = 1, 4/6

hope this helps!

Answer 3

Take the example 3 + 9 + 21 = 33
Divide both sides by 3 getting:
1 +3 + 7 = 11
11= 11
Do you see that you would get an incorrect answer if you did not divide each term by 3?
3 + 9 + 21 = 33
3(1+3+7)= 33 I hope you agree when we factor a polynomial, we take the GCF out of each term, not just selective ones. Well the reverse is also true.

Answer 4

you must divide EVERY TERM on both sides of eq by 3 to get:

x^2 - 5/3x = -2/3 then solve
an easier way is
3x^2 -5x +2 = 0
(3x-2)(x-1)=0
x=1 or x=2/3

Answer 5

Because the 5 is on the same side as the 3x^2. The Fundamental Rule of Algebra says to do the same thing to both SIDES of an equation, not just one term on that side.

Answer 6

Multiply the width by ability of the top, this is the area of a parallelogram. So (3x-4)(x+a million) = 10 improve and rearrange to get a quadratic, the constructive root is the respond. 3x^2 -x -14 = 0, x = 7/3

Answer 7

The whole idea behind the quadratic formula is completing the square. To solve a quadratic equation, you must get everything on one side and set the other equal to zero. Since x is square, you can have up to two real solutions. This is why the quadratic formula works.

ax^2 + bx + c = 0
x^2 + b/a * x = - c/a
x^2 + (b/a)x + b^2/(4a^2) = - c/a + b^2/(4a^2)-->complete the square
(x + b/2a ) ^2 = (-4ac + b^2)/(4a^2)
x + b/(2a) = [ +/- SQRT (b^2-4ac)]/2a
x = -b/2a +/- [SQRT(b^2-4ac)]/2a

If you use the quadratic formula, which is valid as shown, you will get the answer.

Answer 8

When you multiply or divide an equation, you must multiply or divide ALL the terms. Leaving the 5 unchanged "breaks the rules".

Answer 9

Try solving by completing the square, or by using the quadratic formula! It will help you solve this problem quickly!

Here's a link with an example of how to use this formula!

Answer 10

psssh... forget that...

use the quadratic forumula to solve that...

x = -b plus or minus the square root of b squared minus 4ac, divided by 2a....