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secondary 3 | A Maths
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Pls help legit need it by tonight
Date Posted: 3 years ago
Views: 411
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Thanks,but why did the tb only write a=4,b=-40 as the answer.Is it because uf yoy sub in a=-4,b=40 into the equation the equation will not be equivalent
The tutor is incorrect. Or rather, he has failed to check to see if a solution is valid or not.
a = 4 will lead to a negative LHS which cannot be obtained from a square root. This “a = 4” appears in the squaring process as the negative LHS would have been squared to become positive.
a = 4 will lead to a negative LHS which cannot be obtained from a square root. This “a = 4” appears in the squaring process as the negative LHS would have been squared to become positive.
I think you meant a = -4 leads to a negative LHS.
a = 4 means 4√7 - 5 ≈ 5.58, which is positive and therefore equals the square root on the RHS.
@Sunny :
①
For real values to be obtained, the term inside the square root has to be ≥ 0.
And the real value obtained is always ≥ 0.
Eg. √0 = √0 , √81 = 9 , √1024 = 32
②
A square root of a negative number will lead to imaginary numbers , which will only be taught at A levels.
Eg. √-1 = i where i is the imaginary unit.
Eg. √-64 = √64 √-1 = 8i
③
For square roots we always take the principal value which is positive.
Eg. √16 = 4 and not -4 , even though both -4 x -4 = 16 and 4 x 4 = 16
This is why on the RHS , √(137+b√7) will always be ≥ 0 for real values.
Therefore, it cannot equal a negative on the LHS.
So a = -4 is rejected as that would make the LHS negative.
a = 4 means 4√7 - 5 ≈ 5.58, which is positive and therefore equals the square root on the RHS.
@Sunny :
①
For real values to be obtained, the term inside the square root has to be ≥ 0.
And the real value obtained is always ≥ 0.
Eg. √0 = √0 , √81 = 9 , √1024 = 32
②
A square root of a negative number will lead to imaginary numbers , which will only be taught at A levels.
Eg. √-1 = i where i is the imaginary unit.
Eg. √-64 = √64 √-1 = 8i
③
For square roots we always take the principal value which is positive.
Eg. √16 = 4 and not -4 , even though both -4 x -4 = 16 and 4 x 4 = 16
This is why on the RHS , √(137+b√7) will always be ≥ 0 for real values.
Therefore, it cannot equal a negative on the LHS.
So a = -4 is rejected as that would make the LHS negative.
@ Sunny yes that is also correct. So it's important to always check the values to see if they satisfy the original equation..
Thanks guys.My AMath teacher explained this question to me today and he said that if sub in the other a it would not make the equation equivalent so the other a is rejected.He also said that the question has clues that tells you that it is only one answer and that clue would be the question said to find the value and not a pair of values.But thanks for explaining it to so much detail
Yes, your teacher is correct.
The explanation above is actually a logical reasoning such that you can eliminate a = -4 without actually having to calculate the values on both sides for verification.
i.e if a = -4, then a√7 is negative.
So a√7 - 5 would definitely be negative as well since we are subtracting a negative value from another negative value.
It's more of a conceptual thing.
The explanation above is actually a logical reasoning such that you can eliminate a = -4 without actually having to calculate the values on both sides for verification.
i.e if a = -4, then a√7 is negative.
So a√7 - 5 would definitely be negative as well since we are subtracting a negative value from another negative value.
It's more of a conceptual thing.
Okay thanks can you also help me with this question I will post it now
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