quadratic equations

[heatherbird7]

How would I start to work a problem like this, using ac method?
Would I just take the factors of each? I'm a little confused.

6x² – 7x -3 = 0

 

[burakaltr]

How would I start to work a problem like this, using ac method?
Would I just take the factors of each? I'm a little confused.

6x²– 7x -3 = 0

Easy :
if you factor out


6 x**2 -7 x -3
2 -3
* *
* *
3 1


2*(1) + (-3)*(3)=-7

x's coefficient ( -7 ) is satisfied by cross multiplying and then adding

so

( 2*x - 3 ) * ( 3*x +1) is the answer

set to zero

x=3/2

or

x=-1/3

 

[Deleted member 4993]

How would I start to work a problem like this, using ac method?
Would I just take the factors of each? I'm a little confused.

6x²– 7x -3 = 0

You would need to consider factors of [6*(-3)=] -18.

 

[mmm4444bot]

I think that what you're calling the "ac method" I learned as "factor by grouping".

Subhotosh is correct. a*c is -18, so you're looking for two numbers that multiply to make -18 but add to make -7.

Use these two numbers to rewrite the middle term -7x as a sum of two x-terms.

Then factor by grouping.

Please show your work, if you would like more help with this exercise.

 

[heatherbird7]

quadratic equations reply

I'm not really sure that I am confident with what you guys did. I am still unsure of how you came about the factors. I guess the * is a little confusing. Does two * mean raised to whatever power?

 

[heatherbird7]

quadratic equation

That's what I thought originally. Maybe that's why I'm so confused. Maybe I do need to simply factor. Does anyone else oppose?

I think that what you're calling the "ac method" I learned as "factor by grouping".

Subhotosh is correct. a*c is -18, so you're looking for two numbers that multiply to make -18 but add to make -7.

Use these two numbers to rewrite the middle term -7x as a sum of two x-terms.

Then factor by grouping.

Please show your work, if you would like more help with this exercise.

 

[burakaltr]

How would I start to work a problem like this, using ac method?
Would I just take the factors of each? I'm a little confused.

6x²– 7x -3 = 0

Factor 6 the coeff of x**2

Factor -3 the constant term

you can find various combinations. From polynomial equality.

if P(x)=(ax+b)*(cx+d) = a*c*x**2 + (ad+bc)*x + b*d

which says :

find a,c factors of where b,d factors of constant term

and make sure ad+bc = satisfies the coeff of x
like cross-producting

a c

b d

 

[mmm4444bot]

Maybe I do need to simply factor.

Yes, you need to factor the polynomial, so that you can set each factor equal to zero and solve for x.

What you're calling the "ac method" is just one way to factor the polynomial.

Can you answer the following question?

What two numbers multiply to make -18 but add to make -7?

PS: Writing x**2 means x^2 in a computer language called FORTRAN; I don't think that schools teach FORTRAN in algebra classes. You can simply write x^2 to mean x-squared.