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junior college 2 | H2 Maths
3 Answers Below
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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: 272
Good morning Sonia! Doing this question for you now.
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
= (√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
See 3 Answers
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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.
Date Posted:
4 years ago
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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.
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.
Date Posted:
4 years ago
done
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Not sure what kind of algebraic proof is required but this may help.
Date Posted:
4 years ago
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