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junior college 1 | H2 Maths
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How to do part 3 and 4 (iii and iv)?
Date Posted: 4 years ago
Views: 317
iv)
f(2) = (a(2) + b)² = (2a + b)²
When ff(2) = b²,
(a(2a+b)² + b)² = b²
(a(2a + b)² + b)² - b² = 0
(a(2a + b)² + b - b)(a(2a + b)² + b + b) = 0
a(2a + b) = 0 or a(2a + b)² + 2b = 0
a = 0 or b = -2a
or
4a³ + 4a²b + ab² + 2b = 0
ab² + (4a² + 2)b + 4a³ = 0
b = [-(4a² + 2)± √( (4a² + 2)² - 4(a)(4a³) ) ] / 2a
b = [-(4a² + 2) ± √(16a⁴ + 16a² + 4 - 16a⁴)] / 2a
b = [-(4a² + 2) ± √(16a² + 4) ] / 2a
b = [-(4a² + 2) ± 2√(4a² + 1)] / 2a
b = -2a - [1± √(4a² + 1)] / a
f(2) = (a(2) + b)² = (2a + b)²
When ff(2) = b²,
(a(2a+b)² + b)² = b²
(a(2a + b)² + b)² - b² = 0
(a(2a + b)² + b - b)(a(2a + b)² + b + b) = 0
a(2a + b) = 0 or a(2a + b)² + 2b = 0
a = 0 or b = -2a
or
4a³ + 4a²b + ab² + 2b = 0
ab² + (4a² + 2)b + 4a³ = 0
b = [-(4a² + 2)± √( (4a² + 2)² - 4(a)(4a³) ) ] / 2a
b = [-(4a² + 2) ± √(16a⁴ + 16a² + 4 - 16a⁴)] / 2a
b = [-(4a² + 2) ± √(16a² + 4) ] / 2a
b = [-(4a² + 2) ± 2√(4a² + 1)] / 2a
b = -2a - [1± √(4a² + 1)] / a
(^^)
Same answer as answer key?
The answer key says no answer
I think the values that you got should be rejected as a and b are always positive. So there is indeed no answer.
That was my thinking at first too. So yea just reject the answer and quote the relevant reason
See 3 Answers
Sorry to tell you, but you sent your answer to a wrong person.
done
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So the answer is this?
The answer key says no answer
The answer key says no answer
Date Posted:
4 years ago
You should be able to stop at the 3rd step by stating that since a and b > 0, the expression > 0 since all terms of it are positive , therefore no solution
Extra working for extra marks!
Some inaccuracies in the one shown here though. Follow the one I posted
done
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Can expand as given if you are confident but waste time and mistake prone (ie skip let A=(ax+b)^2 and just solve the whole equation).
Can don’t be A if it’s confusing can be C/D/E/F etc.
Can don’t be A if it’s confusing can be C/D/E/F etc.
Date Posted:
4 years ago
So, how did you do iv?
The part 3 I can understand though.
The part 3 I can understand though.
Follow the steps at the bottom of the page where it goes:
“Given ff(x)=b^2” and
“let A=(ax+b)^2”
If still cannot understand, I’ll post the workings.
“Given ff(x)=b^2” and
“let A=(ax+b)^2”
If still cannot understand, I’ll post the workings.
ff(2) actually, not ff(x)
Ah okay my bad
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