I had to solve 2x^2-x-1 using the quad. formula. I got it down to (1 +/- the square root of neg. 8) divided by four. I broke the sq. of neg. 8 into the sq. root of 4 and the sq. root of -2. But my question is could it also be the sq. root of -4 and the sq. root of two? Is one way right or wrong? And when I go from there, I reduced this to 1+/-2 times the sq. root of 2i/4. I further reduced it to 1/4 +/- 1/2 the sq. root of i/2. But I'm not sure if I reduced it properly. Could you show me? Thank you!
Doublecheck the expression within the square root, and watch your signs.
b^2 - 4ac
1 - (4)(1)(-1)
b^2 - 4ac
1 - (4)(2)(-1)
To solve the quadratic equation 2x^2 - x - 1 = 0 using the quadratic formula, you correctly start by plugging the coefficients of the equation into the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -1, and c = -1. Substituting these values into the formula, we have:
x = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2))
x = (1 ± √(1 + 8)) / 4
x = (1 ± √9) / 4
x = (1 ± 3) / 4
Now let's break down the ±3 under the square root. You correctly mentioned two possible ways to do it: √9 = √(3^2) = 3 or √9 = √(4(-2)) = 2i√2.
Using the first option, we have:
x = (1 + 3) / 4 = 4/4 = 1
x = (1 - 3) / 4 = -2/4 = -1/2
So, one solution is x = 1 and the other solution is x = -1/2.
Now, let's recap your further reduction. You correctly expressed the solutions as 1 ± 2√2i / 4. To simplify this fraction, we can divide the numerator and denominator by 2:
x = (1/2) ± √2i / (4/2)
x = (1/2) ± √2i / 2
x = 1/2 ± (1/2)√2i
To make it even more simplified, we can factor out 1/2 from both terms:
x = (1 ± √2i) / 2
Therefore, your final expression should be x = (1 ± √2i) / 2.
To summarize:
The solutions to the quadratic equation 2x^2 - x - 1 = 0 using the quadratic formula are x = (1 + √2i) / 2 and x = (1 - √2i) / 2.