Number of Roots of Polynomial with RadicalDate: 02/05/2002 at 16:55:53 From: Meghan McAdams Subject: Number of roots of polynomial with radical I cannot get a good answer for this problem: Solve this equation: r-9(r^(1/2)) + 8 = 0 factors into: ((r^(1/2) - 8)((r^(1/2)) - 1) r^(1/2) = 8 r = 64 similarly, r = 1 But the highest degree is 1, so therefore why does it have 2 zeros? I know it has something to do with the square root, and then we run into this problem: r-5(r^1/2)-36 factors into: ((r^1/2) - 9) (r^(1/2) + 4) therefore r^1/2 = 9 r = 81 and r^1/2 = -4 There is no solution, since no square root equals a negative number. So this does meet the criteria of having only one root. Therefore, the question is: How do you know how many roots an equation in this form will have? (I know you can look at what the factors will be, but I'm looking more for a theorem or some such.) Thanks for your help and keep up the good work! Meghan Date: 02/05/2002 at 17:11:02 From: Doctor Peterson Subject: Re: Number of roots of polynomial with radical Hi, Meghan. First, your equation is NOT a polynomial, since it includes a radical, so facts about the roots of polynomials do not apply (directly). We can transform it to a polynomial, however, by changing variables. Given r - 9 sqrt(r) + 8 = 0 let s = sqrt(r), so r = s^2; then s^2 - 9s + 8 = 0 (s - 8)(s - 1) = 0 s = 8 or 1 Reversing the substitution, sqrt(r) = 8 or 1 r = 8^2 or 1^2 r = 64 or 1 Thus, we get up to two roots according to the degree of the polynomial; but since the square root disallows negative roots of the quadratic, we may have less than that depending on the specific roots involved. You have to work through the problem to determine how many roots there will actually be. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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