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secondary 3 | A Maths
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Hello everyone! Need help with this please. Is there a solution? And if no why not? Thank you for your guidance! :)
Date Posted: 3 years ago
Views: 200
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x³/² = -20
(x¹/²)³ = -20
(√x)³ = -20
Now, √x ≥ 0 for all real values of x, whereby x ≥ 0
Eg. √4 = 2, √0 = 0, √100 = 10 (we only take the principal value of the square root, we don't consider the negative so we don't say things like √4 = -2 etc.)
√-1 is undefined (actually imaginary , i. But for O levels this is not considered)
So the cube of it will also be ≥ 0
Therefore, (√x)³ ≥ 0 for all real x, x ≥ 0
It is not possible to equal to -20 so there is no solution.
Alternatively,
x³/² = -20
(x³)¹/² = -20
√(x³) = -20
Similarly, for √(x³) to be a real value, x³ ≥ 0
This means that √x³ ≥ 0 so it can never equal -20.
(x¹/²)³ = -20
(√x)³ = -20
Now, √x ≥ 0 for all real values of x, whereby x ≥ 0
Eg. √4 = 2, √0 = 0, √100 = 10 (we only take the principal value of the square root, we don't consider the negative so we don't say things like √4 = -2 etc.)
√-1 is undefined (actually imaginary , i. But for O levels this is not considered)
So the cube of it will also be ≥ 0
Therefore, (√x)³ ≥ 0 for all real x, x ≥ 0
It is not possible to equal to -20 so there is no solution.
Alternatively,
x³/² = -20
(x³)¹/² = -20
√(x³) = -20
Similarly, for √(x³) to be a real value, x³ ≥ 0
This means that √x³ ≥ 0 so it can never equal -20.
Thanks J! I’m curious because if I take square root of a number, it should give me -/+ answer. In this case RHS IS -ve still can’t? Why can’t we square both sides to solve this question? Thanks for your advice!
To answer your question ,
You only do ± IF your number is an unknown.
eg. x² = 4
x = 2 or x = -2
But if you have a KNOWN value , eg. 4
Even though both 2 x 2 = 4 and -2 x -2 = 4,
√4 = 2 and not -2 since we only consider the principal value for square roots of a known value.
Next,
if you were to square both sides ,
x³ = (-20)²
x³ = 400
x = ³√400
But notice what happens when you substitute x = ³√400 into your original equation :
x³/² = (³√400)³/²
= ((400)¹/³)³/²
= (400)¹/²
= √400
= 20
You won't get back -20 , which means x = ³√400 doesn't satisfy the equation.
You only do ± IF your number is an unknown.
eg. x² = 4
x = 2 or x = -2
But if you have a KNOWN value , eg. 4
Even though both 2 x 2 = 4 and -2 x -2 = 4,
√4 = 2 and not -2 since we only consider the principal value for square roots of a known value.
Next,
if you were to square both sides ,
x³ = (-20)²
x³ = 400
x = ³√400
But notice what happens when you substitute x = ³√400 into your original equation :
x³/² = (³√400)³/²
= ((400)¹/³)³/²
= (400)¹/²
= √400
= 20
You won't get back -20 , which means x = ³√400 doesn't satisfy the equation.
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