69: 16x^2+8x-3. I inserted this into the quadratic formula and got -8+/- the square root of (8^2-4(16)(-3)/(2 times 16). This reduced to -8+/- the square root of (64+120)/32, which further reduced to -8+/-the square root of 184/32. I reduced the top part to -8 +/- the square root of four and the square root of six which became -8 +/- 2 times the square root of six over 32. I divided -8 and 32 by 8, resulting in -1 +/- the square root of six/-16. But my book said the answer is 1/4, -3/4. Where did I go wrong?
Thanks for your help!!!
I got
x = (-8 ± √(64 + 192)/32
= (-8 ± √256)/32
= (-8 ± 16)/32
= 8/32 or -24/32
= 1/4 or -3/4
Can you find your error ?
Let's go through the steps together to find out where you made an error. The given quadratic equation is \(16x^2 + 8x - 3 = 0\).
First, let's express the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a = 16\), \(b = 8\), and \(c = -3\).
Now we can substitute the values into the quadratic formula:
\(x = \frac{-8 \pm \sqrt{8^2 - 4(16)(-3)}}{2(16)}\)
Simplifying the equation within the square root:
\(x = \frac{-8 \pm \sqrt{64 + 192}}{32}\)
Continuing to simplify the equation within the square root:
\(x = \frac{-8 \pm \sqrt{256}}{32}\)
Simplifying the square root of 256:
\(x = \frac{-8 \pm 16}{32}\)
Now we have two possible solutions:
\(x_1 = \frac{-8 + 16}{32} = \frac{8}{32} = \frac{1}{4}\)
\(x_2 = \frac{-8 - 16}{32} = \frac{-24}{32} = -\frac{3}{4}\)
So, the solutions to the quadratic equation are indeed \(x = \frac{1}{4}\) and \(x = -\frac{3}{4}\).
Therefore, the book's answer of \(x = \frac{1}{4}\) and \(x = -\frac{3}{4}\) is correct. It appears that you made an error during simplification, likely involving the square root of 256. Double-check your simplification steps to identify the mistake.