Hi, Sam. To solve problem #1, we can use the Pythagorean Theorem. If one leg of the triangle = x, then the other leg = 2x + 1. We know the hypotenuse is 8. So we can use the following formula:
a² + b² = c²
We know that a and b represent the two legs and c is the hypotenuse of the right triangle. We now have:
x² + (2x + 1)² = 8²
x² + 4x² + 4x + 1 = 64
5x² + 4x + 1 = 64
5x² + 4x - 63 = 0
This looks like a good time to use the Quadratic Formula, which says given ax² + bx + c = 0, then:
x = -b + sqrt (b² - 4ac)/2a
I will allow you to substitute into this equation to find the appropriate value(s) for x. Contact me if you need any further assistance with this problem.
2. To solve this problem by completing the square, we first need to divide both sides of the equation by the coefficient of the x² term, which is 6. This gives us:
k² + 17/6 k + 5/6 = 0 Now, we subtract 5/6 from both sides.
k² + 17/6 k = -5/6 The next step of the process is to take 1/2 of the coefficient of the k-term, which is 17/6, and then square that result. One-half of 17/6 is 17/12. Squaring that gives us 289/144. Now, we add 289/144 to both sides of the equation. That yields:
k² + 17/6 k + 289/144 = -5/6 + 289/144 Now, we have a perfect square binomial on the left, and we need to add two fractions on the right-hand side of the equation.
(k + 17/12)² = -5/6 + 289/144 After simplifying the right-hand side, you need to take the square root of both sides and solve for k.
(k + 17/12)² = -120/144 + 289/144 = 169/144
(k + 17/12) = + 13/12
k + 17/12 = 13/12 or k = 13/12 - 17/12 = -4/12 = -1/3 and
k + 17/12 = -13/12 or k = -13/12 - 17/12 = -30/12 = -5/2
So the solutions to this equation are k = -1/3 and k = -5/2
Please contact me if you need any additional assistance. Thank you in advance.
