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So it is true for x = -a. Is it true for all real x? Use algebraic manipulation to simplify the left side to x. bc(a+x)/[(a-b)(a-c)] + ...... = -bc(a+x)(b-c)/[(a-b)(b-c)(c-a)] + ........ etc :) :) :wave:

First a = 2, b = -3 and c = -5 OR a = -2, b = 3 and c = 5 :) :) :wave:

Because 1 + w + w^2 = 0, .: 1 + w = -w^2 :) :) :wave:

If the first equation has real roots, its discriminant >= 0, .: a^2 - b^2 >= 0 , assuming a =/= b (note: I think this is given), then a^2 - b^2 > 0 and the discriminant of the second equation < 0, .: the second equation has unreal roots. If the first equation has unreal roots, its...

Please type your question carefully. Is the first equation 8(a^2)x(2x-1)+b^2=0 ? :) :) :wave:

Given that P(x) = 2x^3 + x^2 - 4x - 2 has a rational zero, instead of using the factor theorem, factorise by grouping the terms appropriately (not always possible) to find the rational zero(s). P(x) = 2x^3 + x^2 - 4x - 2 = (2x^3 + x^2) - (4x + 2) = x^2 ( 2x + 1) - 2(2x + 1) = (2x + 1)(x^2 -...

Not true. There are many ways to detect the motion of the coin from the plane, e.g. you can see the coin becoming smaller as it moves away from you. :) :) :wave:

y=ax^(2n) implies (0,0). Use (-60,18) and (40,8) to set up simul. eqs. 18 = a(-60)^(2n) 8 = a(40)^(2n) 18/8 = (-60/40)^(2n), simplify to 9/4 = (-3/2)^(2n) 9/4 = (9/4)^n .: n = 1 Sub. into 8 = a(40)^(2n) to find a = 1/200 :) :) :wave:

In this case, as the coin travels straight down from your POV you can't tell what state of motion you are in. Only from an outside observer's POV where they saw the parabolic trajectory of the coin can it be said that the plane is moving. The above atatement did not explain why the plane was...

|x+1|/(x^2 - 1) = |x+1|/[(x-1)(x+1)] is defined for all x values except -1 and 1. To simplify the expression, consider the two cases: x < -1 and x > -1. For x < -1, |x+1|/(x^2 - 1) = 1/(1-x) For x > -1, |x+1|/(x^2 - 1) = 1/(x-1). This second case covers x =0, |x+1|/(x^2 - 1) =...

You did not type the formula correctly! :) :) :wave:

You have left out the minus and dx. :) :) :wave:

when |x +1| <0 .......... This is wrong. | | is always greater than or equal to 0, and since x =/= -1, so |x + 1|>0. For x < -1, |x + 1|/[(x - 1)(x + 1)] = -(x +1)/[(x - 1)(x + 1)] = -1/(x - 1)...

Quote 'Correct. Strictly speaking, equations have solutions, while polynomials have roots.' Happy 2006 + 1 :) :) :wave: <!-- google_ad_section_end -->

Dash's investment after 6 years: A = PR^n = 9000(1 + 8/100)^6 =$14281.87 Happy 2006 + 1 :) :) :wave:

P is the initial amount invested. In this question P = 0. Q is amount invested at the end of each year. Superannuation formula. Happy 2007 :) :) :wave:

The equation is correct. dA/dx = 1/2[-x^2 /sqrt(25 - x^2) + sqrt(25 - x^2)] = 0 x^2 = 25/2 so max area = 1/2 (5/sqrt2)(5/sqrt2) = 6.25 Happy 2007 :) :) :wave:

Use A = PR^n + Q(R^n - 1)/(R - 1) to calculate the savings. P = 0, Q = 2000, R = 1 + r/100 = 1.075, n = 5 A = 11616.78 The balance = 15000 - 11616.78 = $3383.22 :) :) :wave:

Prove that the polynomial P(x) = 5x^4 + 2x^3 - 3x^2 - x + 3 does not have integer roots. Possible integer roots are +/- 1, +/- 3. P(+/- 1) =/= 0, P(+/- 3) =/= 0, no integer roots. zeros of a polynomial and roots of a polynomial are the same. Prove that P(x).... cannot have...

Yes, e.g. p(x) = x^2 +3x/2 +1/2, p(-1) = 0, p(-1/2) = 0. q(x) = 2x^2 + 3x + 1, q(-1) = 0, q(-1/2) = 0. q(x) = 2p(x) Note: The factor theorem is used mainly to find rational factors. There are infinite number of polynomials with irrational factors. :) :) :wave: