Question 1329: I have several questions.
1. Find a root of 1/2*x^2 + x + 1 - e^x = 0
2. Find two of the four zeros of x^4 + 2*x^3 - 7*x^2 + 3
3. Find the equation of parabola that passes through (-4,0) and (1,0).
4. Solve for x: cot x cos^2 x = 2 cot x. Give all the solutions.
5. Find the 5th root of 32 (cos 5*(pi)/6 + i sin 5*(pi)/6).
Please help!
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Sol: 1. By the Taylor series of e^x at x=0,we have
e^x = 1 + x + 1/2*x^2 + 1/6 * x^3 + ... = E 1/k! * x^k
(where E means the summation pf k from of all integers k >=0)
Hence, 1/2*x^2 + x + 1 - e^x = -(1/6 * x^3 + 1/24 * x^4+...)
= -x^3(1/6 + 1/24 * x +...) = -x^3(E 1/(k+3)! * x^k)
We see that the final expension isb zero when x=0.
So,the equation 1/2*x^2 + x + 1 - e^x = 0 has a root 0.
2.x^4 + 2*x^3 - 7*x^2 + 3 = 0
First of all ,check that +/-1 or +/- 3 are not roots,so
there is no linear factors of x^4 + 2*x^3 - 7*x^2 + 3.
Try it as the product of two qadratic polynomials as
(x^2 + ax + 1) (x^2 + bx + 3) = x^4+(a+b)x^3 + (ab+4)x^2 +(3a+b)x +3.
Compare the coefficients, we get
a+b = 2..(1)
ab+4 =-7..(2)
3a+b= 0..(3)
From(1),(3) we have a=-1,b=3 but (2) is not true.
Next try x^4 + 2*x^3 - 7*x^2 + 3 =
(x^2 + ax - 1) (x^2 + bx - 3) = x^4+(a+b)x^3 + (ab-4)x^2 -(3a+b)x +3
Compare the coefficients, we get
a+b = 2..(1)
ab-4 =-7..(2)
3a+b= 0, a=-1,b=3
Hence,x^4 + 2*x^3 - 7*x^2 + 3 = (x^2 - x - 1) (x^2 + 3x - 3)
(x^2 - x - 1) = 0 imples x = [1 +/- sqrt(5)]/2
(x^2 + 3x - 3) = 0 imples x = [-3 +/- sqrt(21)]/2
We obtain two zeros as [1 + sqrt(5)]/2 and [-3 + sqrt(21)]/2.
[You can choose any two among the 4 roots as answer.]
3. Since the line passing through (-4,0) and (1,0) is a horizontal line and their midpoint
is (-3/2,0). Assume one possible parabola which vertex is (-3/2, b) and its equation
is y - b = a(x + 3/2)^2.
Set b = 2, and since it passing through (1,0).
Solve -2 = a(1+3/2)^2 = 25a/4, we get a = -8/25.
So,the equation is y - 2 = -8/25(x + 3/2)^2
4. cotx(cos^2 x -2) = 0,
Hence,cot x = 0, by cot x = cos x/sin x, the solutions x: 2n pi +/- pi/2 for integers n.
Or to solve cos^2 x - 2 =0. However, |cos x| <=1, so cos^2 x = 2 has no solutions.
[I feel this question does not make any sense.]
5. Solve z^5 = 32 (cos 5*(pi)/6 + i sin 5*(pi)/6)
= 2^5 (cos [2k pi +5*(pi)/6] + i sin [2k pi + 5*(pi)/6]) for integers k.
By De Meieve Theorem:
z = 2(cos [2 kpi/5 +(pi)/6]+ i sin [2kpi/5+5*(pi)/6]),for k=0,1,2,3,4.
Hence, the 5th roots are 2(cos (pi)/6]+ i sin (pi)/6),
2(cos (17 pi/30)+ i sin (17 pi/30)), 2(cos (29 pi/30)+ i sin (29 pi/30)),
2(cos (41 pi/30)+ i sin (41 pi/30)), 2(cos (53 pi/30)+ i sin (53 pi/30)).
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