Use the quadratic formula to find the real solutions of the equation 2x^2-3x=1
I really need help with this one because I was absent for this portion of the notes because I got called down to guidance can someone please help me and explain this to me
2 x^2 - 3 x - 1 = 0
x = [ 3 +/- sqrt (9+8) ] / 4
= [ 3 +/- sqrt(17) ] / 4
well
if a x^2 + b x + c = 0
then
x = [ -b +/- sqrt(b^2-4ac) ] / 2a
Thank you so very much Damon.
You are welcome.
To find the real solutions of the equation 2x^2 - 3x = 1 using the quadratic formula, you will need to identify the values of the coefficients a, b, and c in the general quadratic equation ax^2 + bx + c = 0.
In this case, a = 2, b = -3, and c = -1.
Now, we can use the quadratic formula, which states that the solutions (x-values) of a quadratic equation can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Let's substitute the values into the formula:
x = (-(-3) ± √((-3)^2 - 4 * 2 * (-1))) / (2 * 2)
Simplifying further:
x = (3 ± √(9 + 8)) / 4
= (3 ± √17) / 4
Hence, the real solutions of the equation 2x^2 - 3x = 1 are (3 + √17) / 4 and (3 - √17) / 4.