Quadratic Help!

[Hossam121]

Hello, this is my first post on this forums!
I'm having trouble with this question: 6x^2-5x-6=0
(3x+2)(2x-3)
Can someone explain to me an easy way on how I can find the 2 numbers (+2 and -3), I know what to do for the rest of it.

Thanks in advance!

 

[Deleted member 4993]

Hello, this is my first post on this forums!
I'm having trouble with this question: 6x^2-5x-6=0
(3x+2)(2x-3)
Can someone explain to me an easy way on how I can find the 2 numbers (+2 and -3), I know what to do for the rest of it.

Thanks in advance!

Do you know the solution for quadratic equation?

If Ax² + Bx + C = 0

then

(x - x₁)(x - x₂) = 0 when

x_1,2 = [-B ± √(B² - 4*A*C)]/(2*A)

Just substitute the values of A, B and C and calculate x₁ and x₂

 

[Hossam121]

Do you know the solution for quadratic equation?

If Ax² + Bx + C = 0

then

(x - x₁)(x - x₂) = 0 when

x_1,2 = [-B ± √(B² - 4*A*C)]/(2*A)

Just substitute the values of A, B and C and calculate x₁ and x₂

Hey, thanks for replying. I do know how to solve a quadratic, just not this particular kind of question.
Could you please just solve 6x^2-5x-6=0 and show me step by step working?

 

[Deleted member 4993]

Hey, thanks for replying. I do know how to solve a quadratic, just not this particular kind of question.
Could you please just solve 6x^2-5x-6=0 and show me step by step working?

Great...

Answer the following questions:

A = ?

B = ?

C = ?

according to your given equation.

 

[stapel]

Hey, thanks for replying. I do know how to solve a quadratic, just not this particular kind of question.
Could you please just solve 6x^2-5x-6=0 and show me step by step working?

Are you saying that you are familiar with the concept of solving by finding the zeroes? And that you are familiar with factoring, but only with "1" as the leading coefficient? And that you're familiar with "a", "b", and "c" as being the labels for the coefficients, but you've never heard of the Quadratic Formula (being the formula the other posters have demonstrated and have asked you to plug into)? Or are you saying something else?

Please be specific. Thank you!

 

[Mrspi]

Hey, thanks for replying. I do know how to solve a quadratic, just not this particular kind of question.
Could you please just solve 6x^2-5x-6=0 and show me step by step working?

You want to factor 6x² - 5x - 6

(in general, this is a trinomial of the form ax² + bx + c)

Step 1: multiply the coefficient of the squared term by the constant term (or, in general, multiply a*c in the above-mentioned general trinomial):

6 * (-6) = -36

Step 2: look for two numbers whose product is what you got in step 1, and whose sum is the coefficient of the middle term, (or b, in the general form). So, we need two numbers whose product is -36 and whose sum is -5. Looks like -9 and +4 will work.

Step 3: Use the two numbers obtained in step 2 to replace the middle term in the original trinomial. We'll use -9x + 4x instead of the original -5x:

6x² - 9x + 4x - 6

Step 4: factor this new version by grouping the first two terms together and the last two terms together:

(6x² - 9x) + (4x - 6)
3x(2x - 3) + 2(2x - 3)
(2x - 3)(3x + 2)

Does that help?

 

[Hossam121]

You want to factor 6x² - 5x - 6

(in general, this is a trinomial of the form ax² + bx + c)

Step 1: multiply the coefficient of the squared term by the constant term (or, in general, multiply a*c in the above-mentioned general trinomial):

6 * (-6) = -36

Step 2: look for two numbers whose product is what you got in step 1, and whose sum is the coefficient of the middle term, (or b, in the general form). So, we need two numbers whose product is -36 and whose sum is -5. Looks like -9 and +4 will work.

Step 3: Use the two numbers obtained in step 2 to replace the middle term in the original trinomial. We'll use -9x + 4x instead of the original -5x:

6x² - 9x + 4x - 6

Step 4: factor this new version by grouping the first two terms together and the last two terms together:

(6x² - 9x) + (4x - 6)
3x(2x - 3) + 2(2x - 3)
(2x - 3)(3x + 2)

Does that help?

Perfect, this is the reply I was looking for!
Thank you very much