SOLUTION: If the two roots of the equation: X^2+X-3=0 are M and N: a. Form the equation that has the roots 1/M, 1/M. b. Form the equation that has the roots M^2, N^2.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: If the two roots of the equation: X^2+X-3=0 are M and N: a. Form the equation that has the roots 1/M, 1/M. b. Form the equation that has the roots M^2, N^2.      Log On


   

Question 871953: If the two roots of the equation: X^2+X-3=0 are M and N:
a. Form the equation that has the roots 1/M, 1/M.
b. Form the equation that has the roots M^2, N^2.
Found 3 solutions by josgarithmetic, mananth, jim_thompson5910:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
(a) in other words, the one root 1/M.
highlight%28%28X-1%2FM%29%5E2=0%29.

(b)
highlight%28%28X-M%5E2%29%28X-N%5E2%29=0%29.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
If the two roots of the equation: X^2+X-3=0 are M and N:
sum of roots = -b/a in equation ax^2+bx +c =0
product of roots = c/a
Therefore in x^2+x-3 =0
M+N = =-1
MN = -3
The roots of the required equation are 1/M & 1/N
First we find the value of 1/M + 1/N and 1/MN and substitute (M+N) and MN
1/M + 1/N = (M+N)/MN = -1/-3 = 1/3
1/M * 1/N = 1/MN = 1/-3= -1/3
The required equation is
x^2-(sum of roots)x + ( product of roots =0
x%5E2-%281%2F3%29x+%2B+%28-1%2F3%29=0
mmultiply equation by 3
3x%5E2-x-1=0

M^ N^2 are the roots
M%5E2%2BN%5E2=+%28M%2BN%29%5E2-+2MN+=++%281%2F3%29%5E2+-+%28-3%29=+1%2F9+%2B3+=+13%2F9
M%5E2N%5E2+=+%28MN%29%5E2+=+%28-3%29%5E2+=+9
frame the equation as done above
x%5E2-+%2813%2F9%29x%2B9+=0
multiply by 9
9x%5E2-13x%2B81+=0


Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

If the two roots of the equation: x^2+x-3=0 are m and n, then

(x-m)(x-n) = x^2+x-3
x^2-xm-xn+mn = x^2+x-3
x^2-(m+n)x+mn = x^2+x-3

Equate coefficients to get
-(m+n) = 1 which leads to m+n = -1
mn = -3

Now square both sides and isolate m^2 + n^2

m+n = -1
(m+n)^2 = (-1)^2
(m+n)^2 = 1
m^2 + 2mn + n^2 = 1
m^2 + 2(-3) + n^2 = 1 ... plug in mn = -3
m^2 - 6 + n^2 = 1
m^2 + n^2 = 1 + 6
m^2 + n^2 = 7

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a) If the roots are 1/m and 1/n, then...

(x - 1/m)(x - 1/n)
x^2 - x/m - x/n + 1/mn
x^2 - (1/m + 1/n)x + 1/mn
x^2 - (n/mn + m/mn)x + 1/mn
x^2 - ((n + m)/mn)x + 1/mn
x^2 - ((m + n)/mn)x + 1/mn
x^2 - ((-1)/mn)x + 1/mn
x^2 - ((-1)/(-3))x + 1/(-3)
x^2 - (1/3)x - 1/3


If the roots are 1/m and 1/n, then the polynomial is x^2 - (1/3)x - 1/3

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b) If the roots are m^2 and n^2, then

(x - m^2)(x - n^2)
x^2 - xn^2 - xm^2 + m^2n^2
x^2 - (n^2 + m^2)x + (mn)^2
x^2 - (m^2 + n^2)x + (mn)^2
x^2 - (7)x + (-3)^2
x^2 - 7x + 9

If the roots are m^2 and n^2, then the polynomial is x^2 - 7x + 9