# How to solve this? (1 Viewer)

#### Cherrybomb56

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

#### Drongoski

##### Well-Known Member
Don't understand your question. What is that small circle??

#### Drongoski

##### Well-Known Member
If question is to solve, say:

$\bg_white \frac {2x(x-4)}{x-1} > 7$

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:
$\bg_white \frac{(2x - 1)(x-7)}{(x-1)} > 0$

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!

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#### foxxraven

##### Member
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

##### Well-Known Member
Or doing the usual way:
$\bg_white \frac {2x(x-4)}{(x-1)} > 7\\ \\ \therefore (x-1)^2\times \left [ \frac {2x(x-4)}{(x-1)}\right ] > (x-1)^2 \times 7 \\ \\ \therefore 2x(x-4)(x-1) > 7(x-1)^2\\ \ \therefore (x-1)[2x(x-4) - 7(x-1) > 0] \\ \\ \therefore (x-1)(2x^2-15x+8) > 0 \\ \\ \therefore (x-1)(2x-1)(x-7) > 0$

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

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#### Cherrybomb56

##### Active Member
Or doing the usual way:
$\bg_white \frac {2x(x-4)}{(x-1)} > 7\\ \\ \therefore (x-1)^2\times \left [ \frac {2x(x-4)}{(x-1)}\right ] > (x-1)^2 \times 7 \\ \\ \therefore 2x(x-4)(x-1) > 7(x-1)^2\\ \ \therefore (x-1)[2x(x-4) - 7(x-1) > 0] \\ \\ \therefore (x-1)(2x^2-15x+8) > 0 \\ \\ \therefore (x-1)(2x-1)(x-7) > 0$

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.

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#### ultra908

##### Active Member
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

##### Active Member
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

##### Well-Known Member
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

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#### Drongoski

##### Well-Known Member
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

##### Well-Known Member
For e):

$\bg_white \therefore sum of roots: \frac {1}{k}\alpha + \alpha + k\alpha = - b \implies (\frac {1}{k} + 1 + k)\alpha = -b \\ \\ sum of root pairs: \frac {1}{k} \alpha \times \alpha + \frac {1}{k} \alpha \times k \alpha + \alpha \times k \alpha = (\frac {1}{k} + 1 + k)\alpha ^2 = c \\ \\ \therefore -b\alpha = c and \alpha = -\frac{c}{b} \\ \\ product of roots: \frac {1}{k} \alpha \times \alpha \times k\alpha =\alpha^3 = -d \\ \\ \implies \left(-\frac{c}{b} \right)^3 = -d \implies -c^3 = -db^3 \implies db^3-c^3 = 0$

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#### Cherrybomb56

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

#### Drongoski

##### Well-Known Member
Cherry

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

#### Cherrybomb56

##### Active Member
Cherry

Finally found my LaTeX error for my solution to e) above.
wow thanks so much.