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secondary 4 | E Maths
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Need help with this as well. I'm not sure how the answer is +_ 1/2 square root m-3
1.You see a square root on the right side. So square both sides of the equation to get rid of the square root first.
t² = (√(4x²/(m-3)))²
t² = 4x²/(m-3)
2. There is a (m-3) in the denominator on the right side. Multiply both sides by (m-3) to get rid of it on the right side.
t²(m-3) = 4x²/(m-3) x (m-3)
t²(m-3) = 4x²
3. There is a 4 on the right side. Divide both sides by 4 to get rid of it.
t²(m-3)/4 = 4x²/4
t²(m-3)/4 = x²
4. Square root both sides of the equation to get rid of the ² on the right side.
± √((t²(m-3))/4) = x
There is a ± because the both the positive form and negative form can give you back x² when they are squared
Now reexpress the square root.
√t²√(m-3) / √4 = x
x = t√(m-3) / 2
x = ½ t √(m - 3)
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