1 math question i can't do?

Question

Solve for x: 12x^2 - 31x + 20 = 0

Answer 1

We need to factor the equation down. This involves finding the roots of the equation. This will look like:

(ax+b)(cx+d) = 12x^2-31x+20 = 0

Because the final term is positive and the middle term is negative, my first hunch is to have b = -4 and d = -5. By doing so, the equation looks like:

(ax-4)(cx-5) = 12x^2-31x+20 = 0

The factors needed for 12x^2 could be 2 and 6 or 3 and 4. We should test each combination to see that they work out. The factoring ends up looking like this:

(3x-4)(4x-5) = 0

Now all that remains is to solve for x in each set of parentheses. If the first term is zero, then

3x-4 = 0
x = 4/3

If the second is zero, then

4x-5 = 0
x = 5/4

Thus, x = 5/4, 4/3

Answer 2

Yeeyy quadratics :d

ok 12x^2 - 31x + 20 = 0

You want to factorise this equation first. Meaning we have to have two brackets with an x term and a number in each

Think of two numbers that multiply to give twelve, and another two numbers that multiply to give 20. One number from each has to multiply with the other number to give two numbers that add or subract to give -31.

Let try 1 and 12 to give 12 and 5 and 4 to give 20

First iam going to try by multiplying the 1 on one side with the 5 on the other, and that leaves the 12 to multiply with the 4

That gives me two numbers which are 5 and (12 x 4) = 48

5 and 48, whatever you do with these two numbers u cant get 31 whether + or -, u just cant get 31 with these numbers.

What if we swapped the two numbers that multiplied by each other, so this time the 1 will multiply by the 4, and the 12 will multiply by the 5
This gives me the numbers 4 and 60. But nope, this didn't work as well. 4 and 60 will never give me 31 as well.

So iam going to pick different numbers to give me 12 and i keep the numbers that give me 20

Let try 3 and 4 to give twelve, and 5 and 4 to give 20 again.
First I am going to try multiplying the 3 with the 5 and the 4 with the other 4.

This gives me two numbers that are 15 and 16. Yep, these two numbers DO give us 31 either by 15 + 16, or -15 -16 = 31 = -31.
But since we want -31, we want both terms minus.

Now the important part is that it is only the numbers we took at the beginning which we will put in the brackets, not the 15 and 16. The 3 and 4 for the x terms (becuase 12x^2) and the 5 and 4 for the number 20.

So making two brackets with these numbers, the 3x must multiply with the -5 and the 4x must multiply with the -4 to give -15x and -16x which will simplify to -31x.The 3x multiplies with 4x to give 12x^2 and -4 multiplies with -5 to give 20.
So we have
(3x - 4)(4x -5) = 0
Notice it's important to get the signs the right way round when you do these.

With these, multiplying everything inside the brackets with everything outside gives us the equation we started with, so this is correct.

Now for something to equal to 0 with two numbers multiplying, one of the numbers must equal to zero.

So of these brackets one or the other must equal to zero, becuase they multiply each other.

Lets do the first bracket equals 0

3x - 4 = 0
3x = 4
x = 4/3<< this is one answer for x

Now the other bracket

4x -5 = 0
4x = 5
x = 5/4<< this is the second answer for x

It is quite common to have two answers for x in a quadratic equation to give 0. So the two values for x in this equation to give 0 is 4/3 and 5/4 (keeping them as fractions).

If you want to check if this is correct subsitute these numbers one by one into the equation and see if in both cases they give zero, which they do. So were correct ;)

Sorry for the long complicating explanation for the factorising bit, theres a way on paper which makes this method very simple but since I am not on paper thats the only way I could explain. Try seperating the two numbers that give 12 and the others that give 20 by a thick line.

Answer 3

Here's help with the algebra homework.

12x^2 - 31x + 20 = 0
(4x-5)(3x-4) = 0
x= 5/4 or 4/3

I haven't done algebra like that since high school (over 10 years), can't belive I can still do it.

Answer 4

12x^2-31x+20 = 0 first factor
(3x-4)(4x-5) = 0 now solve each for x

3x-4 = 0
x = 4/3

4x-5 = 0
x = 5/4 , so

x = 4/3 and x = 5/4

Answer 5

use the quadratic equation where

a= 12, b=31 c=20

Answer 6

Try the formula: -b+- square root of (bsquared minus 4ac) all over 2a. This is the "quadratic formula".

Using this formula gives two roots of the given quadratic equation, which are: 4/3 and 5/4 or 1.333.... and 1.25. These are the two values of x which solve the equation.

Answer 7

Use the quadratic equation:

http://mathworld.wolfram.com/QuadraticEquation.html

I get x= 4/3
x=5/4

Answer 8

A good first step if you can't factor it easily is to use the quadratic equation since it works for all cases.

x = -b (+or-) sqrt(b^2-4ac)/2a

where ax^2 + bx +c = 0

Answer 9

12x^2-31x= -20
It is the answer b/c it can't be simplified anymore.

Answer 10

(1.35,x) is the only x-intercept, or what x is equivilent to. Graphing such functions can be gone on a calculator, or http://www.shodor.org/interactivate/activities/GraphSketcher/