# roots of the equation

• Oct 15th 2009, 02:12 PM
scrible
roots of the equation
I have been doing good so far with these problems but I have hit a road block and just can't seem to get pass it. The problems is as follows:
equation is 2x^2-4x+5=0 and the roots are a and b. I am suppose to finde the value of 1/(a+1) + 1/(b+1). I have already fond that a+b= -b/a = -(-4/2)= 2 and ab=c/a=5/2. Is there anyone out there who can help me?(Worried)
• Oct 15th 2009, 04:34 PM
skeeter
Quote:

Originally Posted by scrible
I have been doing good so far with these problems but I have hit a road block and just can't seem to get pass it. The problems is as follows:
equation is 2x^2-4x+5=0 and the roots are a and b. I am suppose to finde the value of 1/(a+1) + 1/(b+1). I have already fond that a+b= -b/a = -(-4/2)= 2 and ab=c/a=5/2. Is there anyone out there who can help me?(Worried)

$\frac{1}{a+1} + \frac{1}{b+1}$

$\frac{(b+1) + (a+1)}{(a+1)(b+1)}$

$\frac{a+b+2}{ab + a + b + 1}$

finish now?
• Oct 15th 2009, 05:58 PM
pacman
2x^2 - 4x + 5 = 0, dividing BOTH SIDES by 2, we have x^2 - 2x + 5/2 = 0

a + b = 2 and ab = 5/2

Or as suggested by SKEETER: evaluate this http://www.mathhelpforum.com/math-he...8537f4e0-1.gif . . . .

http://www.mathhelpforum.com/math-he...8537f4e0-1.gif = [(a + b) + 2]/[(ab) + (a + b) + 1] = [2 + 2]/[5/2 + 2 + 1] = 4/[5/2 + 3] = 4/[11/2] = 8/11.

Or THIS,

x^2 - 2x + 5/2 = 0, let x = y - 1, so that the root will be x + 1

(y - 1)^2 - 2(y - 1) + 5/2 = 0,

y^2 - 2y + 1 - 2y + 2 + 5/2 = 0,

y^2 - 4y + 3 + 5/2 = 0

y^2 - 4y + 11/2 = 0.

let y = 1/z, so that the root will be 1/(x + 1),

(1/z)^2 - 4(1/z) + (11/2) = 0, multiply both sides by (2/11)z^2,

2/11 - (8/11)z + z^2 = 0, re-arranging,

z^2 - (8/11)z + 2/11 = 0,

notice that the sum of roots is 8/11. (Bow)

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