Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Help please
Date Posted: 3 years ago
Views: 580
See 1 Answer
√(5x - 3) = -√(4y + 1)
For √(5x - 3) to be a real value, 5x - 3 ≥ 0
(The term inside the Square Root has to be 0 or positive real value)
5x ≥ 3
x ≥ 3/5
This means that ① √(5x - 3) ≥ 0 when x ≥ 3/5 (a positive real square root is always ≥ 0)
Likewise, in order for √(4y + 1) to be a real value, 4y + 1 ≥ 0
4y ≤ -1
y ≤ -¼
This means that √(4y + 1) ≥ 0 when y ≤ -¼
Then ② -√(4y + 1) ≤ 0 when y ≤ -¼ (a negative real square root ≤ 0)
Comparing ① and ②, the former is 0 or positive but the latter is 0 or negative.
Now there is no way a positive valie equals a negative value. The only way to satisfy √(5x - 3) = -√(4y + 1) is when both sides equal 0.
i.e x = 3/5, y = -¼
Therefore there is only one solution
For √(5x - 3) to be a real value, 5x - 3 ≥ 0
(The term inside the Square Root has to be 0 or positive real value)
5x ≥ 3
x ≥ 3/5
This means that ① √(5x - 3) ≥ 0 when x ≥ 3/5 (a positive real square root is always ≥ 0)
Likewise, in order for √(4y + 1) to be a real value, 4y + 1 ≥ 0
4y ≤ -1
y ≤ -¼
This means that √(4y + 1) ≥ 0 when y ≤ -¼
Then ② -√(4y + 1) ≤ 0 when y ≤ -¼ (a negative real square root ≤ 0)
Comparing ① and ②, the former is 0 or positive but the latter is 0 or negative.
Now there is no way a positive valie equals a negative value. The only way to satisfy √(5x - 3) = -√(4y + 1) is when both sides equal 0.
i.e x = 3/5, y = -¼
Therefore there is only one solution
It is very easy to disprove the result given in the question.
Eg.
5x = 4y + 4
For √(5x - 3) to be a real value, x ≥ 3/5
If x = 8, then 5(8) = 4y + 4
40 = 4y + 4
4y = 36
y = 9
So x = 8 when y = 9
But when x = 8, √(5x - 3) = √(5(8) - 3) = √37
Yet, when y = 9, -√(4y + 1) = -√(4(9) + 1) = -√37 ≠ √37
So x = 8, y = 9 does not satisfy the original equation.
Squaring both sides has eliminated the negative aspect of the RHS in the original equation (after being rewritten as √(5x - 3) = -√(4y + 1) )
So for √(5x - 3) + √(4y + 1) = 0 to be true for all real values of x and y, both square root can only equal 0
i.e x = 3/5, y = -¼ is the only real valued solution.
Eg.
5x = 4y + 4
For √(5x - 3) to be a real value, x ≥ 3/5
If x = 8, then 5(8) = 4y + 4
40 = 4y + 4
4y = 36
y = 9
So x = 8 when y = 9
But when x = 8, √(5x - 3) = √(5(8) - 3) = √37
Yet, when y = 9, -√(4y + 1) = -√(4(9) + 1) = -√37 ≠ √37
So x = 8, y = 9 does not satisfy the original equation.
Squaring both sides has eliminated the negative aspect of the RHS in the original equation (after being rewritten as √(5x - 3) = -√(4y + 1) )
So for √(5x - 3) + √(4y + 1) = 0 to be true for all real values of x and y, both square root can only equal 0
i.e x = 3/5, y = -¼ is the only real valued solution.
360 Tutoring Program