How to solve this? (1 Viewer)

[Cherrybomb56]

Does anyone know how to solve this inequality? I don't quite get how to approach it.

 

[Drongoski]

Don't understand your question. What is that small circle??

 

[Drongoski]

If question is to solve, say:



can do this in a number of ways. One is using sign diagrams. Note because of the denominator, note x must not equal 1.

The above inequality is equivalent to:


3 relevant factors are: 2x-1, x-7 and x-1 which are respectively equal to zero at x = 0.5, 7 and 1. Mark these numbers on a number line. Test for the sign of the expression at a convenient point in one of the 4 intervals defined by the 3 points; say in the interval x <= 0.5, choose, say x = 0 (or -1, -2 or -3 or whatever number less than half). The expression becomes (2*0-1)(0-7)/(0-1) = (-ve)(-ve)/(-ve) = -ve. So the 1st interval x < 0.5 is -ve. Then the successive 3 open intervals to its right are +ve(0.5<x<1), -ve(1<x<7) and +ve(x > 7)); i.e. the signs alternate: -|+|-|+

So reading off this sign diagram: answer is 0.5 < x < 1 or x > 7.


Not sure if I'm right. Have to double check!

 

[foxxraven]

Do you access your textbook through Box of Books? I have Maths in Focus as well and my school uses BoB and the digital copy does that weird circle thing with some of the inequality signs. Check the question in your physical book.

 

[Drongoski]

Or doing the usual way:


You can then do a rough sketch of this cubic polynomial and read off the positive intervals.

 

[Cherrybomb56]

Or doing the usual way:


You can then do a rough sketch of this cubic polynomial and read off the positive intervals.

thanks for the help, i understood it well. if you could can you also help me in these three questions as I am not understanding how to do them.

 

[ultra908]

1. if (x-1) divides the polynomial, and it divides (x^2-1)Q(x), it must divide the remainder (kx+2). Thus you can figure out k. Then, using this, divide the polynomial by (x+1). Since (x+1) already divides (x^2-1)Q(x), the remainder is found when (kx+2) is divided by (x+1)

2. Using the remainder theorem, the remainder when divided by (x-1) is equivalent to P(1), since if P(x)=(x-1)Q(x) + k, P(1) = k. Similarly, dividing by x gives the remainder P(0). this should give u the equation to find k.

3. Using the relationship between roots and coefficients, find b c and d in terms of a and k. Sub into equation to show it equals 0.

 

[Cherrybomb56]

1. if (x-1) divides the polynomial, and it divides (x^2-1)Q(x), it must divide the remainder (kx+2). Thus you can figure out k. Then, using this, divide the polynomial by (x+1). Since (x+1) already divides (x^2-1)Q(x), the remainder is found when (kx+2) is divided by (x+1)

2. Using the remainder theorem, the remainder when divided by (x-1) is equivalent to P(1), since if P(x)=(x-1)Q(x) + k, P(1) = k. Similarly, dividing by x gives the remainder P(0). this should give u the equation to find k.

3. Using the relationship between roots and coefficients, find b c and d in terms of a and k. Sub into equation to show it equals 0.

Could you please explain question 1 further as I am not quite understanding what you mean.

 

[Drongoski]

Cherry: I know you want a very gentle step-by-step help. Not helpful hints here and there.

For b) P(x) = (x^2-1)Q(x) + kx + 2
If x-1 is a factor of P(x), then the Remainder Thm tells you that P(1) = 0
i.e. P(1) = (1^2 - 1)Q(1) + k + 2 = 0 ==> 0 + k + 2 = 0 so that k = -2.

You now know that P(x) = (x^2-1)Q(x) - 2x + 2
For remainder from dividing by x+1, we just find P(-1) = (1-1)Q(-1) -2*(-1) + 2 = 0 + 2 + 2 = 4

 

[Drongoski]

For c):

P(x) = x^3 + 4x^2 + 2x + k

You are told: remainder upon division by (x-1) = 2 times remainder upon division by x or (x-0). That means

P(1) = 2 P(0)

i.e. 1 + 4 + 2 + k = 2 x k
.: k = 7

 

[Drongoski]

For e):

 

[Cherrybomb56]

Thank you so much for your help. you guys truely are amazing and have explained it so well.

 

[Drongoski]

Cherry

Finally found my LaTeX error for my solution to e) above.

 

[Cherrybomb56]

Cherry

Finally found my LaTeX error for my solution to e) above.

wow thanks so much.