Quadratic Functions (1 Viewer)

[Real Madrid]

1
____ <1

x-2

Solve for x.

Tell me the easiest method for finding out how to solve this inequation.

 

[bawd]

Multiply both sides by the square of the denominator.
(x-2)^2/(x-2) < (x-2)^2

Simplify LHS.
(x-2) < (x-2)^2

Then move (x-2)^2 over to the left.
(x-2) - (x-2)^2 < 0

Factorise out (x-2)
(x-2)(1-(x-2)) < 0

Simplify
(x-2)(3-x) < 0

x-2 < 0 and 3-x <0

Therefore x < 2 or x > 3

 

[Real Madrid]

I write up new formulas and steps to solve problems, usually for parabola it is:

FIND
1. whether a>0,a<0
2.Use -b/2a and sub the result to find y
3.Find y intercept
4. Find x intercept(s).

For x^2-5x+6, i get the point (2.5,-4) as my turning point.

graphimatica says otherwise that its (2.5,-0.25)?

What am I doing wrong?

edit: ignore this, forgot to square x

 

[kurt.physics]

There are 3 ways of doing this problem.

1) Multiplying by the square of the denominator

2) Determining when the expression is positive and negative and just solving the two cases

3) Sketching the function y = 1/(x-2) and y = 1, then the solution is where the graph is below the line y = 1.

If you would like me to solve these different ways just tell me.

 

[bored of sc]

The denominator of the algebraic fraction with ALWAYS be a factor when you solve it. So if you get to the factored part of your solution and can't see it you've done something wrong.

 

[Real Madrid]

Well what's the quickest way to solve such problems and get full marks?

 

[bored of sc]

Well what's the quickest way to solve such problems and get full marks?

I only know one way.

Multiply by denominator squared. Come up with solutions to x (treat like an equation i.e. change < or > to =).

Substitute values of x into original equation to work out where the answer lies.

In your example:

1
____ <1

x-2

Solve for x.

Multiply both sides by (x-2)² and change < to =.

(x-2)²/(x-2) = (x-2)²

Cancel down and expand where appropriate.

x-2 = x²-4x+4

Rearrange and solve quadratic equation.

x²-5x+6 = 0

By factorisation:
(x-2)(x-3) = 0

x = 2, 3

Substitute x = 0 into original equation:

-1/2 < 1 True

Therefore x < 2, x > 3.

If that helps, cool.

 

[Real Madrid]

I'll use your method then>

I just find that inequalities are my weak point in maths.

It's simple but sometimes i just crack

 

[tommykins]

Uh, the quickest way to do it

1/(x-2) < 1
x-2 < (x-2)²
x-2 < x² - 4x + 4
0 < x²-5x + 6
0 < (x-3)(x-2)

Draw the graph, for whatever values the graph is positive (as it is 0<) that's the value for x.
Also watch out for <= symbols, make sure your denominator =/= 0.

 

[PC]

I actually prefer a different method. Treat the question in "cases".

Case 1: The denominator is positive.
So if x – 2 > 0, we have x > 2
And therefore:
1 / (x – 2) < 1
Multiply by the denominator (which is positive):
1 < x – 2
Add 2 to both sides
3 < x
x > 3
Since x > 2 and x > 3, we can see that x > 3.

Case 2: The denominator is negative.
So if x – 2 < 0, we have x < 2
And therefore:
1 / (x – 2) < 1
Multiply by the denominator (which is negative), so the inequality needs to be reversed:
1 > x – 2
Add 2 to both sides
3 > x
x < 3
Since x < 2 and x < 3, we can see that x < 2.

So the full solution is x < 2 and x > 3.

The same method works for absolute value questions too.

 

[Continuum]

Uhh... that takes ages PC...

 

[lyounamu]

Uhh... that takes ages PC...

True...but when the really complicated question arises, that method is more useful.

With the first one, it's really difficult to use when the number gets too complicated.

 

[bored of sc]

Uh, the quickest way to do it

1/(x-2) < 1
x-2 < (x-2)²
x-2 < x² - 4x + 4
0 < x²-5x + 6
0 < (x-3)(x-2)

Draw the graph, for whatever values the graph is positive (as it is 0<) that's the value for x.
Also watch out for <= symbols, make sure your denominator =/= 0.

Why not just treat like an equation and sub in a values for x into the original equation to work out if it is above or below your two answers? That saves the time of doing a graph.

 

[untouchablecuz]

1
____ < 1

x-2

x =/= 2

Now treat the inequality sign as an '='.

1 = x-2

x = 3

Now, visualize a number line:

--------2------3---------

Test any number, 0 for example:

1
____ < 1 -0.5 < 1

(0)-2

. ' . x<2, x>3

LOL, I know that's a crap explanation, but thats how i do em, transforming them into quadratics takes much longer and is more error prone.

 

[lyounamu]

1
____ < 1

x-2

x =/= 2

Now treat the inequality sign as an '='.

1 = x-2

x = 3

Now, visualize a number line:

--------2------3---------

Test any number, 0 for example:

1
____ < 1 -0.5 < 1

(0)-2

. ' . x<2, x>3

LOL, I know that's a crap explanation, but thats how i do em, transforming them into quadratics takes much longer and is more error prone.

I like your method. That was the method that my teacher taught us the first time.

 

[tommykins]

Why not just treat like an equation and sub in a values for x into the original equation to work out if it is above or below your two answers? That saves the time of doing a graph.

The graph takes 2 seconds to draw whereas subbing in 2 values will take time (especially if you have fractions/decimals).

 

[bored of sc]

The graph takes 2 seconds to draw whereas subbing in 2 values will take time (especially if you have fractions/decimals).

Fair enough.

 

[jellybelly59]

hey ppl how do u do...:
15/(x-2) ≤ x

 

[tommykins]

15(x-2) ≤ x(x-2)²
0 ≤ x(x-2)² - 15(x-2)
0 ≤ (x-2)[(x(x-2) - 15]
0 ≤ (x-2)[ x² - 2x - 15 ]
0 ≤ (x-2)(x-5)(x+3)

Drawing the graph, it is positive at -
-3 ≤ x ≤ 2
and 5 ≤ x

But x =/= 2

Thus -3 ≤ x < 2 and 5 ≤ x

 

[lyounamu]

hey ppl how do u do...:
15/(x-2) ≤ x

15(x-2) <= x(x-2)^2
15(x-2) - x(x-2)^2 <= 0
(x-2)(15-x(x-2)) <=0
(x-2)(15+2x - x^2) <= 0
(x-2)(x-5)(x+3) <=0
Here just use the quadratic formula to find all the points where the graph intersects the x-asxis. Then check to see,