Question

Consider the equation: \[2x^2-20x=x^2-19\]

1) rewrite the equation by completing the square. your equation should look like \[(x c)^2=d\] or \[(x-c)^2=d\].
2) what are the solutions to the equation?
choose 1 answer:
(choice a) \[x=-10 \pm 81\] a \[x=-10 \pm 81\]
(choice b) \[x=10 \pm 81\] b \[x=10 \pm 81\]
(choice c) \[x=-10 \pm 9\] c \[x=-10 \pm 9\]
(choice d) \[x=10 \pm 9\] d \[x=10 \pm 9\]

Answer 1

To rewrite the equation, we add the square of half the coefficient of the x term to both sides, resulting in (x - 10)^2 = x^2 + 81. The solutions to the equation are x = 10 ± 9.

Step-by-step explanation:

To rewrite the equation by completing the square, we first move all terms to one side of the equation:

2x^2 - 20x = x^2 - 19

Next, we add the square of half the coefficient of the x term to both sides:

2x^2 - 20x + (-20/2)^2 = x^2 - 19 + (-20/2)^2

This simplifies to:

(x - 10)^2 = x^2 - 19 + 100

(x - 10)^2 = x^2 + 81

Now we can see that c = 10 and d = 81.

To find the solutions to the equation, we take the square root of both sides:

x - 10 = ±√(x^2 + 81)

x = 10 ± √(x^2 + 81)

Therefore, the correct solution is (choice d) x = 10 ± 9.