SOLUTION: Evaluate 2√3x+1+3√4x+5=25

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Question 1206732: Evaluate 2√3x+1+3√4x+5=25

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
assuming you have:

2sqrt%283x%2B1%29%2B3sqrt%284x%2B5%29=25....square both sides

%282sqrt%283x%2B1%29%2B3sqrt%284x%2B5%29%29%5E2=25%5E2



4%283x%2B1%29%2B12sqrt%283x+%2B+1%29+%2Asqrt%284x+%2B+5%29%2B9%284x%2B5%29=625

12sqrt%283x+%2B+1%29+%2Asqrt%284x+%2B+5%29=625-4%283x%2B1%29-9%284x%2B5%29

12sqrt%283x+%2B+1%29+%2Asqrt%284x+%2B+5%29=576+-+48x....square both sides

%2812sqrt%283x+%2B+1%29+%2Asqrt%284x+%2B+5%29%29%5E2=%28576+-+48x%29%5E2

144%283x+%2B+1%29+%2A%284x+%2B+5%29=2304x%5E2+-+55296x+%2B+331776

1728+x%5E2+%2B+2736+x+%2B+720=2304x%5E2+-+55296x+%2B+331776

2304x%5E2+-+55296x+%2B+331776-1728+x%5E2+-+2736+x+-+720=0

576x%5E2+-+58032x+%2B+331056=0


using quadratic formula we get

x+=+403%2F8+%2B+%2825+sqrt%28201%29%29%2F8
x+=+403%2F8+-+%2825+sqrt%28201%29%29%2F8

verify solutions:
=> false

=> true


so, solution to 2sqrt%283x%2B1%29%2B3sqrt%284x%2B5%29=25 is

x+=+403%2F8+-+%2825+sqrt%28201%29%29%2F8





Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


Just some comments about the problem....

Presumably, this problem is an exercise for the student in solving equations involving the sum of two square root expressions. The process for solving that kind of equation is straightforward -- square both sides of the equation, isolate the remaining square root expression, and square both sides again, resulting in a quadratic equation that can be solved by factoring or by using the quadratic formula.

If that is the case, then the solution should be a "nice" number; and the expressions in the calculations on the way to the answer should be relatively nice.

With this problem, the calculations on the way to the answer are not nice....

If we start from the position that we assume the answer is a nice number, we can see that the sum of the two radical expressions is a whole number, so it would be nice if the two radical expressions both simplified to whole numbers.

If we then look at possible whole number values for x which make both radical expressions whole numbers, we see we want (3x+1) and (4x+5) to both be perfect squares. A quick bit of playing with numbers shows that x=5 does that.

But then the sum of the two radical expressions is 23, not 25.

So my suspicion is that the right side of the equation is supposed to be 23.

But the work to get that solution is still rather ugly. And solving the problem by the standard method leads to a quadratic equation that is virtually impossible to solve by factoring, which means you need to use either the quadratic formula or a tool like a graphing calculator. And if that is the case, then you might as well solve the original problem using a graphing calculator.

So....

(1) My guess is that the right side of the equation is supposed to be 23 instead of 25
(2) Even with that change, I think the problem is probably not appropriate for a learning exercise.