Use the quadratic formula to solve each of the following quadratic equations...
1. 2x^2-5x=3
2. 3x^2-2x+1=0
Rearrange the equation in quadratic formula form.
2x^2 -5x -3 = 0
Then use the formula. Tell me what you don't understand about it if you don't know what to do next.
I am just not good in math at all and I was needing assistance, I am not sure on how to even start it...
We provide assistance in the form of hints to help you get started. But we leave the smaller details to you. We don't DO the homework but we help you do it. To do the first one, make the equation, as I suggested above, read
2x2 -5x -3 = 0.
Remember the quadratic formula is
x = [-b +/- sqrt (b^2 - 4ac)]/2a
where a is the coefficient of the x^2 term, b is the coefficient of the x term and c is the constant. In this case, a is 2, b is -5 and c is -3. Plug those values into the quadratic formula and solve for x
To solve the first equation, 2x^2 - 5x - 3 = 0, we can use the quadratic formula. The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a represents the coefficient of the x^2 term, b represents the coefficient of the x term, and c represents the constant term.
Let's plug in the values for this equation:
a = 2
b = -5
c = -3
Now we can substitute these values into the quadratic formula:
x = (-(-5) ± √((-5)^2 - 4 * 2 * (-3))) / (2 * 2)
Simplifying further:
x = (5 ± √(25 + 24)) / 4
Combining like terms:
x = (5 ± √49) / 4
The square root of 49 is 7, so we get:
x = (5 ± 7) / 4
This gives us two possible solutions:
x1 = (5 + 7) / 4 = 12 / 4 = 3
x2 = (5 - 7) / 4 = -2 / 4 = -0.5
Therefore, the solutions to the equation 2x^2 - 5x - 3 = 0 are x = 3 and x ≈ -0.5.
Now let's move on to the second equation:
3x^2 - 2x + 1 = 0
Following the same steps, we can use the quadratic formula:
a = 3
b = -2
c = 1
Substituting these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4 * 3 * 1)) / (2 * 3)
Simplifying further:
x = (2 ± √(4 - 12)) / 6
Simplifying the radical:
x = (2 ± √(-8)) / 6
At this point, we can see that the expression under the square root, -8, is negative. This means that the equation has no real solutions. The solutions would be complex numbers.