How do you solve using the completing the square method #x^2 - 14x = 0#?

Question

Isaiah Anthony · Accepted Answer

#x# is #0, 14# We begin with #x^2-14x=0#. I like to deal with equations in standard form (#color(red)(a)color(green)(x)^2+color(blue)(b)color(green)(x)+color(orange)(c)#), so I'll just rewrite this to #x^2-14x+0=0#. From here, we need to solve for #x#, and we need to do that by completing the square. Completing the square is a method where we take an equation that is not a perfect square and find a value that could make it factorable. For us, our first step is to make sure that the #color(red)(a)# in #color(red)(a)color(green)(x)^2+color(blue)(b)color(green)(x)+color(orange)(c)# is a #1#. Is that the case for #x^2-14x+0=0#? Yes, it is. Now we take the second coefficient (#color(blue)(b)#), and divide it in half (#-14/2#), in our case giving us #-7#. We then take that value and square it, giving us #49# (#-7^2#). We take that number and add it to our equation, like this: #x^2-14x+49+0=0#. WAIT!!!! We just added a random number into this equation! We can't do that. We need to keep the equation equal. We could add a #49# on the other side, or we could just subtract the #49#. Then the numbers cancel each other out and don't change the value of the equation. NOW we have #x^2-14x+49-49+0=0#. Now, we went to a lot of effort to get #x^2-14x+49#, because this is a perfect square. We can rewrite that part of the equation into #(x-7)^2#. We still need to deal with #-49+0#, so we just combine them to give us #-49#. Put it together and we get #(x-7)^2-49=0#. Now we just need to solve for #x#. I'll go through this part quickly. #(x-7)^2-49=0# #(x-7)^2=49# #sqrt((x-7)^2)=sqrt(49)# #x-7=+-7# #x=7+-7# #x=0, 14#

Savannah Anderson · Answer

undefined To solve the equation $x^2 - 14x = 0$ using the completing the square method:

1. Move the constant term (in this case, 0) to the other side of the equation.
2. Factor out the coefficient of the $x^2$ term, if necessary.
3. Add and subtract the square of half the coefficient of the $x$ term to both sides of the equation to complete the square.
4. Simplify and solve for $x$.
5. Check the solutions for accuracy.

Applying these steps:

1. Move the constant term to the other side: $x^2 - 14x = 0 \rightarrow x^2 - 14x + 49 = 49$
2. Factor out the coefficient of $x^2$: $x^2 - 14x + 49 = (x - 7)^2 = 49$
3. Add and subtract the square of half the coefficient of $x$: $(x - 7)^2 - 49 = 0$
4. Solve for $x$: $x - 7 = \pm \sqrt{49} \rightarrow x - 7 = \pm 7$
5. Determine the solutions: $x = 7 \pm 7$ which gives $x = 14$ and $x = 0$.