The lengths of the sides in a right triangle form three consecutive terms of a geometric sequence. Find the common ratio of the sequence. There are two distinct answers.
A. 10.0 cm B. 34.6 cm C. 85.1 cm D. 120.0 cm which on would it be ?>
i need to find the common ratio not a length
Ok. so let the required ratio be R. Then is it true that - the sides will be a, aR, aR^2 with aR^2 as the largest side i.e. hypotenuse?
i had a/r,a,ar but sure
@luckythebest
then is it true that - a^2 + a^2R^2 = a^2R^4 By pythagorean theorem?
yup
soooo
is it further true that - a^2(1+R^2) = a^2R4 => 1+R^2 = R^4 ?
Let the side lengths be \(1, a, a^2\) for some nonnegative number \(a\). If \(a > 1\), then \(a^2\) is the longest side so \((a^2)^2=(1)^2+(a)^2\). If \(0 < a < 1\), then \(1\) is the longest side so \((1)^2=(a)^2+(a^2)^2\). We can omit the \(a=1\) case since that would be an equilateral triangle. This should give you enough information to come up with the two distinct answers.
@yakeyglee who said 0<a<1
and yes lucky
Next - Let R^2 = K Then is it true that - 1+K = K^2 ?
yup
We have - K^2 - K - 1 = 0 Compare with ax^2 + bx + c = 0 a = 1, b = -1, c = -1 Find b^2-4ac and substitute it in the quadratic formula. Solve for K, and tell me the answer.
If we assume \(0<a<1\) that is one of the two cases. It affects how you treat the computation.
@yakeyglee if you substitute ^ in the quadratic formula you anyway get 2 distinct answers; example x+a and x-a, isn't it?
(1(+-)sqrt5)/2
Ok. so K = 1 +/- Sqrt 5/2 But we told earlier that K = R^2 Hence R^2 = 1 +/- sqrt 5/2 R = Sqrt of R^2. The answer is a bit complex but that is how its done. :(
so r=sqrt((1+/- sqrt5)/2)
:like:
?
i mean its correcto :)
k thanks
If you want further simplification sqrt 5 = 2.236 then R = (1+2.236/1.414) => sqrt of 3.236 / 1.414 (1.414 is sqrt of 2)
@luckythebest I didn't use any x in my computation... i don't understand what you are talking about. And you will get multiple values for \(a\), but I bet only one of them satisfies the initial assumption that was made to derive the equation.
@yakeyglee what i meant to say was that if you use the quadratic equation formula anyway you will get two distinct answers.
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