Using factoring to solve each quadriatic equation. 2x^2-5x-3=0.
This is what I have so far but don't know what to do from here.
GCF:1
A:2
B:-5
C:-3
AC:-6
Goal is the find two numbers that add to -5 and multiply to -6
Factors of 6: 1,6
-2,3 (That's the one)
I don't know what to do from here thanks for your help!!
(2x+1)(x-3)=0
x= -1/2, or x=3
You must be learning a method called "decomposition"
So you want two numbers
whose sum is -5
and whose product is -6
your choice of -2,3 does not work,
but -6,1 does.
So we are going to replace the middle term -5x with
-6x + x
2x^2-5x-3=0
2x^2 - 6x + x - 3 = 0
now group it using common factor in pairs
2x(x-3) + 1(x-3)
common factor again
(x-3)(2x+1) = 0
I am sure you can finish the solving part ...
If you can find a pair of numbers which satisfy the sum and product condition, this method will always work.
To solve the quadratic equation 2x^2 - 5x - 3 = 0 using factoring, you've already taken the correct initial step by finding the GCF (Greatest Common Factor) and identifying the values of A, B, and C.
Now, as you mentioned, the next goal is to find two numbers that add up to -5 and multiply to -6. This is because the quadratic equation can be factored into two binomials in the form: (px + q)(rx + s) = 0, where p and r are the coefficients of x^2, q and s are the coefficients of x, and pq = AC.
In this case, A = 2, B = -5, and C = -3. The product of A and C is AC = 2 * -3 = -6.
Since you've correctly determined that -2 and 3 are the two numbers that add up to -5 and multiply to -6, you can now express the quadratic equation as:
(2x + 3)(x - 1) = 0
To find the solutions, you set each binomial equal to zero and solve for x:
2x + 3 = 0
x - 1 = 0
For the first equation, you subtract 3 from both sides:
2x = -3
Divide both sides by 2:
x = -3/2 or -1.5
For the second equation, you add 1 to both sides:
x = 1
So the complete solutions to the quadratic equation 2x^2 - 5x - 3 = 0 are x = -3/2, -1.5, and x = 1.