Can you solve for X? | (Right triangle) | #math #maths | #geometry

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Can you solve for X? | (Right triangle) | #math #maths | #geometry

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[Skank_and_Gutterboy]

I used a slightly different method and after substitution got the polynomial:
u^3 + u^2 - u = 0, because I pressed on without factoring out and cancelling 5^2x. Factoring out a u: u(u^2 + u -1) = 0. u=0 is eliminated as a solution because, when back-substituting and calculating x, the equation contains the expression "log(0)". log(0) is undefined, so there's no valid x.

For u = (-sqrt(5)-1)/2, that yields a complex value for x. That x satisfies the equation derived using the Pythagorean Theorem but when calculating and checking side lengths of the triangle (5^x, 25^x, and 125^x), all of the triangle sides are negative or complex. That makes no physical sense, so that x is discarded.

Therefore, the only valid solution for x is x ≈ -0.149. My method was a little different but it all worked out in the end. Hoo yah!!! ❤👍

[michaeldoerr]

I incorrectly calculated the golden answer. And I am glad that I have check my work with that video. And the answer is log_5(sqrt(sqrt(5)-1/2))

[jimlocke]

Solving with a python computer program doing a half interval search: for any value of x, the difference between the sum of the squares of the sides and the square of the hypotenuse represents an error. If x = 0, the difference is positive and if x = -0.5, the difference is negative. Since the difference increases monotonically, there is one crossing point where the error is 0. Let y = 2x. Then the error for y = -1 is negative and for y = 0 is positive. Try y = -0.5. If we get a positive result, we next try -0.75, half way between -1 and -0.5. If we get a negative result, we next try -0.25, half way between -0.5 and 0. In this way, we keep halving the distance to 0 error. We quit when we reach the desired precision.

# let y = 2x
y = -0.5
dy = 0.25


# define function to compute sum of squares of the sides minus square of the hypotenuse

def compute_delta(y):
delta = (125.0)**y + (25.0)**y - (5.0)**y
return(delta)

# note that delta for y = -1.0 is negative and for y = 0.0 is positive
# therefore, there is a zero crossing for delta between y = -1.0 and y = 0.0
# if a negative value is returned for y, we must increase y and try again
# if a positive value is returned for y, we must decrease y and try again
# start in the middle of the range, y = 0.5
# once the increment dy is sufficiently small, compute and print x

for a in range(0, 40): # after 40 divisions, dy is below 10 raised to -12
delta = compute_delta(y)
if (delta > 0.0):
y = y - dy
else:
y = y + dy
dy = dy/2
x = y/2
x = round(x, 9) # round x to 9 decimal places
formatted = "{:.9f}".format(x)
print(f"x = ", formatted)

Program output:

x = -0.149496859

=== Code Execution Successful ===

Comparing with RAG981's solution, there is a difference of 1 in the ninth decimal place.

[zdrastvutye]

there 3 ways to solve this:
way 1 and two:
10 print "premath-can you solve for x":sw=.1:x=-5:goto 50
20
50 gosub 20
60 dg1=dg:x1=x:x=x+sw:x2=x:gosub 20:if dg1*dg>0 then 60
70 x=(x1+x2)/2:gosub 20:if dg1*dg>0 then x1=x else x2=x
80 if abs(dg)>1E-10 then 70
90 xl1=x/2:print "loesung 1";xl1:x=1:goto 130:rem newtonsches verfahren
100
110
120 dx=zx/nx:x=x-dx:return
130 gosub 100:if abs(dx)<1E-10 then else 130
140 print "loesung
150 print "der fehler=";fe;"%"
premath-can you solve for x
loesung 1-0.149496859
loesung 2:x=-0.149496859
der fehler=-7.66807841E-11%
>
run in bbc basic sdl and hit ctrl tab to copy from the results window.

way 3:

substitution x^2-x^4-x^6=0
=x*x*(1-x^2-x^4)
3
11% 11% fehler=3.14807588E-9%
>
z=sqr(5)/2-.5:x=ln(z)/2/ln(5)
print z, x

0.618033989
-0.149496859

[yalchingedikgedik]

Thanks Sir
Thanks PreMath
That’s very useful to learn exponential exercises.
With my respects
❤❤❤❤

[MdArbaz]

5^2x=25^2x+125^2x
Or, (5^2)^x=(5^2)^2x+5^2x*25^2x
Or, (5^2)^x=(5^2)^2x+(5^2)^x*(5^2)^2x
Or, 1=(5^2)^x+(5^2)^2x
U^2+U-1 = 0

[JSSTyger]

Assuming only a real solution I get x = ln((-1+sqrt(5))/2)/(2ln(5))

[shapoursefidi]

Why you did not divide 5 by 5 and will be 1 that equation is wrong and is not equal .

[joeschmo]

Yeh, once I saw that the hypotenuse was the smallest of the numbers, I knew x would be negative. Kept going to the point where it started getting messy with the square roots and wouldn't just be a nice easy x=-2 or whatever, so I gave up and let The Calculator People finish it. 😂🧠💯🥳😂

[santoshjha]

Thanks
Can we say pythgoras as Bodhayan sulva sutra?

[prossvay]

5^x=(25)^x+(125)^x
5^2x=(5^2x)2+(5^3x)2
5^2x=(5^2x)^2+(5^2x)^3
Let t=5^2x (t>0)
t=t^2+t^3
t(t^2+t-1)=0
t=0 rejected
t^2+t-1=0
So t=(√5-1)/2 ; t=(-1-√5)/2rejected
So t=(√5-1)/2
So 5^2x=(√5-1)/2
2x=log (5^(√5-1)/2
So x=1/2log (5^(√5-1)/2.❤❤❤

[RAG]

Mine gives -0.149496858. You approximated a term in the calculation and then used it to produce the final answer incorrectly. OOps.
I used x= (log(rt5-1)-log2)/ 2log5. You can use any kind of logarithm, but I prefer the ones that are on my calculator.

[gaskellr]

Fairlt tedious maths problem really, bit of a damp squib.

[LuisdeBritoCamacho]

(5^X)^2 = (25^X)^2 + (125^X)^2

5^(2X) = 25^(2X) + 125^(2X)

5^(2X) = 5^(4X) + 5^(6X)

1 = 5^(2X) + 5^(4X)

5^X = Y

1 = Y^2 + Y^4

Y = +/- 0, 78615

5^X = +/- 0, 78615

ln 5^X = ln 0, 78615

X = ln 0, 78615 / ln 5

X ~ 0, 149498

[Birol]

My way of solution ▶
In the given triangle Δ(ABC):
[AB]= 125ˣ
125= 5³
⇒
[AB]= 5³ˣ

[BC]= 25ˣ
25= 5²
⇒
[BC]= 5²ˣ
If we apply the Pythagorean theorem we get:
[AB]²+[BC]²= [CA]²
[AB]= 5³ˣ
[BC]= 5²ˣ
[CA]= 5ˣ
⇒
(5³ˣ)²+(5²ˣ)²= (5ˣ)²
5⁶ˣ + 5⁴ˣ = 5²ˣ
Let's say:
5²ˣ= u
⇒
u³+u²= u
u³+u²-u=0
u(u²+u-1)=0
⇒
u= 0 ❌
we know that u > 0

u²+u-1=0
Δ= 5
u₁= (-1+√5)/2


u₂= (-1-√5)/2 ❌
u₂ < 0
⇒
u= u₁
u₁= (√5-1)/2
⇒
5²ˣ= (√5-1)/2
5ˣ = √[(√5-1)/2]
if we take ln both sides, we get:
xln(5)= ln [√[(√5-1)/2]]
x= ln [√[(√5-1)/2]]/ln(5)
x= -0, 24060.../1, 609437
x= - 0, 14949