1. Using analytical method, solve the inequality x^2 - 4x + 5 <0
I solved this question by completing the sqyare and got (x-2)^2 + 1 <0
I was stuck after that and there was this "alternative method" provided for us. It says," Since the coefficient of x^2 is positive and discriminant, D = (-4)^2 - 4(5) < 0, therefore, teh graph of y = x^2 - 4x + 5 is above the x-axis, i.e. x^2 - 4x + 5>0 for all real vlaues of x. Hence x^2 - 4x + 5 < 0 has no solution.
I don't get why x^2 - 4x + 5>0. As seen by my solution using completing the square, I followed the initial sign and still got a '<' sign. Anyway, what do i do after completing the square?
2. Using an algebraic method, solve the inequality x^2 + 3x - 1 > 0 (solve this using completing the square, stated by question)
I got (x + 1.5)^2 - 13/4 > 0
Subsequently, i do not know what steps to take to solve the question (find the value of x)
Thanks
Originally posted by anpanman:1. Using analytical method, solve the inequality x^2 - 4x + 5 <0
I solved this question by completing the sqyare and got (x-2)^2 + 1 <0
I was stuck after that and there was this "alternative method" provided for us. It says," Since the coefficient of x^2 is positive and discriminant, D = (-4)^2 - 4(5) < 0, therefore, teh graph of y = x^2 - 4x + 5 is above the x-axis, i.e. x^2 - 4x + 5>0 for all real vlaues of x. Hence x^2 - 4x + 5 < 0 has no solution.
I don't get why x^2 - 4x + 5>0. As seen by my solution using completing the square, I followed the initial sign and still got a '<' sign. Anyway, what do i do after completing the square?
2. Using an algebraic method, solve the inequality x^2 + 3x - 1 > 0 (solve this using completing the square, stated by question)
I got (x + 1.5)^2 - 13/4 > 0
Subsequently, i do not know what steps to take to solve the question (find the value of x)
Thanks
Hi anpanman,
For 1), notice that (x-2)^2 + 1 will always be more or equal to 1, since (x-2)^2 will always be more or equal to 0. Thus, the inequality x^2 - 4x + 5 < 0 has no solution.
When you are using the "completing the square" method, you should not bring the < 0 in for this case.
For 2), From (x+(3/2))^2 - 13/4 > 0, bring 13/4 to the other side to get (x+(3/2))^2 > 13/4. Square root both side, and you will realise that x + 3/2 must be less than [- (sqrt 13/4)] or more than (sqrt 13/4). I believe you should be able to solve this from here.
Cheers.
Hi,
It is possible to have cases where the solution set is empty (as in Q1) or the solution set is the entire real set, say
x^2 + 1 > 0.
Thanks.
Cheers,
Wen Shih
Thanks Mr Wee and Trueheart.
By the way, I do have another question.
Solve the inequality 1/ (1+x)^2 more than or = 1
My answer is -2 less than or equal to x < -1 OR -1 < x less than or equal to 0.
Not sure if the answer is right but it differs slightly form the answer given and I have been trying to spot error in my workings, to no avail
Originally posted by anpanman:Thanks Mr Wee and Trueheart.
By the way, I do have another question.
Solve the inequality 1/ (1+x)^2 more than or = 1
My answer is -2 less than or equal to x < -1 OR -1 < x less than or equal to 0.
Not sure if the answer is right but it differs slightly form the answer given and I have been trying to spot error in my workings, to no avail
Hi anpanman,
Notice that (1+x)^2 will always be more or equal to zero, and thus you can multiply it on both sides to get 1 > or = to (1+x)^2 without any changes to the inequality.
However, the first thing to take note of is that x cannot be -1, as 1/0 is undefined.
From 1 > or = to (1+x)^2, you will have sqrt 1 > or = to 1+x. Hence, 1+x lies between -1 and 1, inclusively. From here, you should be able to continue with the final step.
Cheers.
Hey guys, I also have a question pertaining to inequality. LOL. hahaha.
Solve the inequality (x-1)/(2x-1) < or = x / (4-3x)
I did the necessary and got (-5x^2 + 8x - 4)(-3x+4)(2x-1) < or = 0
2 questions.
a) When I draw the graph to represent the curve, do I need draw y=(-5x^2 + 8x - 4)(-3x+4)(2x-1) or y = (3x+4)(2x-1) since we know the 1st eqn has D <0? In other words, do we ignore (-5x^2 + 8x - 4)?
b) Do we needa state that (-5x^2 + 8x - 4) has a -ve x^2 coefficient and D<0 and thus no solution can be obtained from that part? (not sure about this since (-5x^2 + 8x - 4) is a sub-equation and not just a sole equation alone)
thank you
Originally posted by TrueHeart:Hi anpanman,
Notice that (1+x)^2 will always be more or equal to zero, and thus you can multiply it on both sides to get 1 > or = to (1+x)^2 without any changes to the inequality.
However, the first thing to take note of is that x cannot be -1, as 1/0 is undefined.
From 1 > or = to (1+x)^2, you will have sqrt 1 > or = to 1+x. Hence, 1+x lies between -1 and 1, inclusively. From here, you should be able to continue with the final step.
Cheers.
Hello Trueheart, but I thought we're supposed to multiply the "square" of the denominator to both sides? which means you'll get (1+x)^2 more or = (1+x)^4...
Anyway, i could do with a lil bit of clarification on this part too, hahahah.
Originally posted by Audi:Hello Trueheart, but I thought we're supposed to multiply the "square" of the denominator to both sides? which means you'll get (1+x)^2 more or = (1+x)^4...
Anyway, i could do with a lil bit of clarification on this part too, hahahah.
Hi,
If the expression of the denominator is always positive, we do not need to multiply both sides by its square. Thanks.
Cheers,
Wen Shih
Originally posted by Audi:Hey guys, I also have a question pertaining to inequality. LOL. hahaha.
Solve the inequality (x-1)/(2x-1) < or = x / (4-3x)
I did the necessary and got (-5x^2 + 8x - 4)(-3x+4)(2x-1) < or = 0
2 questions.
a) When I draw the graph to represent the curve, do I need draw y=(-5x^2 + 8x - 4)(-3x+4)(2x-1) or y = (3x+4)(2x-1) since we know the 1st eqn has D <0? In other words, do we ignore (-5x^2 + 8x - 4)?
b) Do we needa state that (-5x^2 + 8x - 4) has a -ve x^2 coefficient and D<0 and thus no solution can be obtained from that part? (not sure about this since (-5x^2 + 8x - 4) is a sub-equation and not just a sole equation alone)
thank you
Hi,
(b) -5 x^2 + 8x - 4 is always negative, by what you have stated.
(a) Because of (b), (-3x + 4)(2x - 1) >= 0, from which you may obtain the set of values of x.
Also, take note of the fact that x = 1/2 and x = 4/3 are inadmissable.
Thanks.
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
(b) -5 x^2 + 8x - 4 is always negative, by what you have stated.
(a) Because of (b), (-3x + 4)(2x - 1) >= 0, from which you may obtain the set of values of x.
Also, take note of the fact that x = 1/2 and x = 4/3 are inadmissable.
Thanks.
Cheers,
Wen Shih
Thanks.
But still, which graph do I plot? y=(-5x^2 + 8x - 4)(-3x+4)(2x-1) or y = (3x+4)(2x-1)? I realized, that even if we draw y=(-5x^2 + 8x - 4)(-3x+4)(2x-1), the GC does not include the part on (-5x^2 + 8x - 4) because it shows a quadratic curve.
Hi,
Just sketch a quick graph (no need to use GC) of y = (-3x + 4)(2x - 1) and obtain the set of values of x for which the curve lies above the x-axis. Thanks.
Cheers,
Wen Shih