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junior college 2 | H2 Maths
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Sonia
Sonia

junior college 2 chevron_right H2 Maths chevron_right Singapore

Hello! Please help me with this qns for both parts:-)

I can’t do the first part when i replace x with u^2 and root x with u ..
for eg. 5-4rootx+x =0
5-4u+u^2=0

*no solution*

so i don’t know how to continue.

Thanks so much

Date Posted: 4 years ago
Views: 242
Eric Nicholas K
Eric Nicholas K
4 years ago
Good morning Sonia! Doing this question for you now.
J
J
4 years ago
5 - 4√x + x

= (√x)² - 2(2√x) + 2² + 1

= (√x - 2)² + 1


For all x ≥ 0, √x is a real number and √x ≥ 0

(√x - 2)² also ≥ 0 since √x - 2 is also a real number and the square of a real number is always 0 or positive.

So (√x - 2)² + 1 ≥ 1 > 0

Which means that (√x - 1)² > 0

Thus 5 - 4√x + x > 0 for all x ≥ 0

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
A very good morning to you Sonia! Here are my workings for the algebraic portion of this question. I send you the remaining part in 1 min, since I cannot fit everything into a page.
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Good morning Sonia! This is an extension of what you have learnt in quadratic inequalities two years ago.

In fact, this approach I have done now can be used for quadratic inequalities as well! Except that you learnt to draw a quadratic curve for the quadratic inequality when you learnt them two years ago.

Here, we must be careful with the inequality as the presence of the square root means that x cannot be negative as well.

Sonia, let me know if you need more explanation.
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Lim En Jie
Lim En Jie's answer
73 answers (Tutor Details)
Not sure what kind of algebraic proof is required but this may help.