<p>Hey guys. I just have a question about the bluebook's page 361, #8 question. how do you get the answer?</p>
<p>the question is:</p>
<p>square root of (x^2-t^2)=2t-x</p>
<p>If 'x' and 't' are positive numbers that satisfy the equation above, what is the value of x/t?</p>
<p>the answer here btw is 5/4 or 1.25</p>
<p>i just cant seem to get the answer.</p>
<p>please help. thanks in advance!</p>
<p>anyone please?</p>
<p>Get rid of the square root:</p>
<p>(original equation)</p>
<p><=> x^2 - t^2 = (2t - x)^2
<=> x^2 - t^2 = 4(t^2) - 4tx + x^2
<=> -5(t^2) = -4tx
<=> 5t = 4x
<=> x/t = 5/4</p>
<p>DinhNyugen,</p>
<pre><code> Square root of x^2 - t^2 = (2t - x)
</code></pre>
<p>Square both sides: x^2 - t^2 = (2t - x)^2
FOIL the RIGHT side: x^2 - t^2 = (4t^2 - 2tx - 2tx + x^2)
Combine like terms: x^2 - t^2 = (4t^2 - 4tx + x^2)
Subtract x^2 from both sides: - t^2 = 4t^2 - 4tx
Subtract 4t^2: -5t^2 = -4tx
Divide by t: -5t = -4x
Divide by -4: (5/4)t = x
Divide by t: 5/4 = x/t</p>
<p>^ haha I wanted to give a detailed explanation but I’m not used to describe math problem solving. </p>
<p>and… you wrote my name wrong :(</p>
<p>thanks for the quick answer guys!</p>
<p>DinhNguyen,</p>
<p>(1) I actually addressed the message to you by accident instead of wakekeboy - whoops! I wasn’t trying to correct what you did I just wrote the wrong name - I was just offering an explanation to wakekeboy</p>
<p>(2) Simply an accidental typo with the name - apologies</p>
<p>I missed the square root component at first.</p>
<p>Anyway, I’ll offer my work, just because.
sqrt(x^2 – t^2) = 2t – x
x^2 – t^2 = 4t^2 – 2tx – 2tx + x^2
x^2 – t^2 = 4t^2 – 4tx + x^2
-t^2 = 4t^2 = -4tx
-5t^2 = -4tx
-5t = -4x
x = 5
t = 4</p>
<p>^ x is not necessarily 5 and t is not necessarily 4; 5/4 is just the ratio of x to t.</p>
<p>For example x, y could be 20, 16; 20/16 = 5/4</p>
<p>Right. For the sake of this problem, though, my solution is fine. Thanks for pointing that out.</p>