@@SyberMath Ugly. . You had to consider 2 semicircles -one with rad 12 and center (0,0) and second with rad 5 and center (0,13). Then your right triangle will appear naturally but not from nowhere. That approach would be educative: if you see sqrt(a-b^2) think about circles.
Nice, and the general solution to sqrt(a^2 - x^2) = c - sqrt(b^2 - x^2) is x = a*b*sin(theta)/c, where theta is the internal angle between sides a and b of the triangle with sides a, b and c.
sqrt(144 - x^2) = 13 - sqrt(25 - x^2) Let a = sqrt(144 - x^2) and b = sqrt(25 - x^2). So, we have a = 13 - b, or a+b = 13. Next, note that a^2 = 144 - x^2 and b^2 = 25 - x^2. Subtracting those, we get a^2 - b^2 = 119. But a^2 - b^2 = (a+b)(a-b). So, (a+b)(a-b) = 119. And we know that a+b = 13, so a-b = 119/13. Then a+b - (a-b) = 13 - 119/13 = 169/13 - 119/13 = 50/13. But a+b - (a-b) = 2b. So, 2b = 50/13, or b = 25/13. Substituting back in, we get sqrt(25 - x^2) = 25/13. Squaring both sides, we get 25 - x^2 = 625/169, so 25 - 625/169 = x^2, or 4225/169 - 625/129 = x^2, or 3600/169 = x^2. So x = +/- 60/13
The whole 25/4 multiplication/division thing has helped me out a lot of times. That and 50/2. And I figured it had something to do with Pythagoras, what with all the 12s, 5s, and 13s, since that's a Pythagorean triple.
The geometric solution is awesome. Just have to be careful it only provides the positive solution. The original question is algebraic and the negative solution also satisfy it.
I feel the substitution is not really necessary here. If you just leave x^2 you can solve it the same way like the first method ending up with an easy quadratic equation, which you do anyway when you back substitute.
I let A=144 - x^2 B=25 - x^2. Then we have the system sqrt(A) +sqrt(B) =13 A - B =119. Square the first equation A + B + 2*sqrt(AB)=169 add this to A - B =119 and 2*A= 288 - 2*sqrt(AB). Divide by 2: A=144 - sqrt(AB). Substitute B=A-119 A=144 - sqrt(A*(A-119)). sqrt(A*(A-119)) =144 - A. Now square both sides: A*(A-119) = 144^2 -288A +A^2. and A^2 cancel. Simplifying 169A = 12^4 A=12^4 / 13^2 x^2 = 144 -A = 12^2 - (12^4 )/ (13^2) =12^2 * (1 - (12^2) /(13^2)) = (12^2)*(5^2 ) / (13^2 ) x=60/13 x=-60/13
Nice, but I immediately recognized the geometric interpretation as bprp already did a similar problem a few months ago: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Z5gwIIX4zdo.html He used a (3, 4, 5) triangle whereas you have a (5, 12, 13) triangle. Of course the Pythagorean triplets are a dead giveaway. If you do want to solve this equation algebraically, there is a better way to do it than your first method. If we rewrite the qiven equation as (1) √(144 − x²) + √(25 − x²) = 13 and then multiply both sides by √(144 − x²) − √(25 − x²) we get (144 − x²) − (25 − x²) = 13(√(144 − x²) − √(25 − x²)) which gives (2) √(144 − x²) − √(25 − x²) = 119/13 and adding (1) and (2) we have 2√(144 − x²) = 13 + 119/13 = 169/13 + 119/13 = 288/13 so (3) √(144 − x²) = 144/13 Squaring both sides of (3) we get 144 − x² = 144²/13² or 12² − x² = 12⁴/13² so x² = (12²·13²)/13² − 12⁴/13² = (12²·(13² − 12²))/13² = (12²·5²)/13² which gives x = 60/13 ⋁ x = −60/13 In general, if we have √(a² − x²) + √(b² − x²) = √(a² + b²) then x² = a²b²/(a² + b²) and since we then also have √(a² − x²) = a²/√(a² + b²) √(b² − x²) = b²/√(a² + b²) it follows that we also have x² = √(a² − x²)·√(b² − x²) This is of course in agreement with the geometric interpretation. If we have a right angled triangle with rectangular sides with lengths a and b and its hypotenuse has a length c and h is the length of the altitude on the hypotenuse, then h = ab/c. In some high school books this is written as ba = ch and therefore inappropriately called the Bach theorem. The altitude divides the hypotenuse in two pieces with lengths √(a² − h²) = p and √(b² − h²) = q and indeed we also have h² = pq.
I used the first method, but without substitution. The second method is beautiful, I must say, but of course you only get x=60/13 since there's no such thing as negative distance.
