Good problem, in that it illustrates a situation wherein one of the two solutions to a quadratic equation is not a solution to the original equation, even though the two equations are linked.
😮 la raiz cuadrada de de 5x debe ser mayor o igual a cero, asi que hacemos un cambio de variable t= ✓5x y la ecuación se convierte en t^2 +5t- 50 =0 de donde se obtiene t1=-10 y t2=5 y de aqui se deduce que t1= -10 no es solucion por ser un valor menor a 0
Can be solved even simpler: √(5x) + x = 10, x ≥ 0 (√x + √5/2)² - 5/4 = 10 (√x + √5/2)² = 45/4 √x + √5/2 = 3√5/2 (only the positive value is valid) √x = √5 x = 5.
Your work is wrong. 5x must have grouping symbols around it. 5 × 5 must have grouping symbols around it. Sqrt(5x) + x = 10, etc. Sqrt(5 × 5) + 5 = 10, etc.
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Good problem, in that it illustrates a situation wherein one of the two solutions to a quadratic equation is not a solution to the original equation, even though the two equations are linked.
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at 0:25 √(5x) = 10 - x, => domain of x is 0 ≤ x ≤ 10, so x = 20 is discarded immediately.
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@@ScholarTutors No. I had studied math at university about 50 years ago.
Wow, thanks
20 is a valid solution; the square root of 100 is ±10, so sqr(100) + x is 30 for the positive, or 10 for the negative.
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NB: your is wrong
😮 la raiz cuadrada de de 5x debe ser mayor o igual a cero, asi que hacemos un cambio de variable t= ✓5x y la ecuación se convierte en t^2 +5t- 50 =0 de donde se obtiene t1=-10 y t2=5 y de aqui se deduce que t1= -10 no es solucion por ser un valor menor a 0
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😊😅х=5 моментально ❗
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*@ ScholarTutors* -- You did not write the radical symbol all the way over "5x,: at least in the first two lines. That is unclear and sloppy.
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√(5x) = 10 - x 5x = 100 - 20x + x² x² - 25x + 100 = 0
(x - 5)(x - 20) = 0 x = 20 is rejected , thus the correct answer is x = 5
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Can be solved even simpler:
√(5x) + x = 10, x ≥ 0
(√x + √5/2)² - 5/4 = 10
(√x + √5/2)² = 45/4
√x + √5/2 = 3√5/2 (only the positive value is valid)
√x = √5
x = 5.
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@@ScholarTutors No, but maths is a big part of my field (EE).
X=5
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@@ScholarTutors SANTO DOMINGO DOMINICAN REPÚBLIC
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√(5*5)+5=√(5*5)+5.
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Nice Olympiad Exponential Equation: √(5x) + x = 10; x =?
10 > x > √(5x) > 0; √(5x) + x = 10 = 5 + 5 = √[5(5)] + 5; x = 5
Answer check:
x = 5: √(5x) + x = 10; Confirmed as shown
Final answer:
x = 5
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instead of formula use factoring method:
x^2-25×+100=(×-20)(×-5)=0
×-20=0, ×=20
×-5=0, x=5
×1=20. ×2=5
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@@ScholarTutors In this simple example, you can apply the factoring method. But with more complex problems, you cannot easily find the factors.
Do not use the times sign for the x-variable.
√(5x) + x = 10
√5 * √x + (√x)² = 10
substitute √x = t
√5 * t + t² = 10 |-10
t² + √5 * t - 10 = 0
t₁,₂ = -√5/2 ± √( (√5/2)² - (-10) )
t₁,₂ = -√5/2 ± √( 5/4 + 40/4 )
t₁,₂ = -√5/2 ± √(45/4)
t₁,₂ = -√5/2 ± √(9 * 5)/2
t₁,₂ = -√5/2 ± 3√5/2
t₁ = -√5/2 + 3√5/2
t₁ = 2√5/2
t₁ = √5
t₂ = -√5/2 - 3√5/2
t₂ = -4√5/2
t₂ = -2√5
resubstitute t₁:
√x₁ = √5 |²
x₁ = 5
test x₁:
√(5x) + x = 10
√(5 * 5) + 5 = 10
√(5²) + 5 = 10
5 + 5 = 10 → correct
resubstitute t₂:
√x₂ = -2√5 |²
x₂ = (-2)² * 5
x₂ = 4 * 5
x₂ = 20
test x₂:
√(5x) + x = 10
√(5 * 20) + 20 = 10
√100 + 20 = 10
10 + 20 = 10 → incorrect, due to the fact, that we artificially raised √x to the power of 2, creating an extraneous solution.
If this solution had turned the leading 10 into -10 (which is impossible, due to the root), it would have been correct.
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√5x+x=10
√5x=10-x
5x=100-20x+x^2
0=100-25x+x^2
x^2-25x+100=0
(x-5)(x-20)=0
x=5,20
↓
x=5
√5×5+5=10
x=20
√5×20+20≠10
↓
Answer x=5
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見たのは0:00から0:08までです
Your work is wrong. 5x must have grouping symbols around it. 5 × 5 must have grouping symbols around it. Sqrt(5x) + x = 10, etc.
Sqrt(5 × 5) + 5 = 10, etc.