Question - Quadratic Formula and Coefficients
Solution:
The equation in the image appears to be the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For a quadratic equation in the form $$ax^2 + bx + c = 0$$. However, it appears that there is a missing term in the denominator (it should be $$2a$$, but only $$2(1)$$ is written), and the coefficients 'a', 'b', and 'c' are not specified in the image. The formula shown is:\[x = \frac{-1 \pm \sqrt{1^2 - 4(8)(20)}}{2(1)}\]This suggests 'a' = 8, 'b' = 1, and 'c' = 20 for the quadratic equation. However, given the typical structure of a quadratic equation $$ax^2 + bx + c = 0$$, it's unusual to have an 'a' term larger than 'b' or 'c'. It's possible there's a mistake in the handwriting. Assuming the 'a', 'b', and 'c' values as 1, 1, and -160 respectively (if we take the coefficient of 'a' to be 1 instead of 8), the equation would be:\[x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-160)}}{2(1)}\]Now let's solve it with these coefficients:\[x = \frac{-1 \pm \sqrt{1 + 640}}{2}\]\[x = \frac{-1 \pm \sqrt{641}}{2}\]We now have two possible solutions for x, which are:\[x_1 = \frac{-1 + \sqrt{641}}{2}\]\[x_2 = \frac{-1 - \sqrt{641}}{2}\] These are the two solutions to the quadratic equation based on the given formula and assuming 'a' = 1. If 'a' is actually 8 as shown, then the quadratic equation would be incorrect, and we would need the correct coefficients to proceed.
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