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.Com
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)   (Show Source): You can put this solution on YOUR website!
(a) in other words, the one root 1/M.
.

(b)
.

Answer by mananth(16946)   (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

mmultiply equation by 3


M^ N^2 are the roots


frame the equation as done above

multiply by 9



Answer by jim_thompson5910(35256)   (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

--------------------------------------------------------------------------------------------
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

--------------------------------------------------------------------------------------------
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

RELATED QUESTIONS

If the two roots of the quadratic equation x^2+x-3=0 are M and N. Form the quadratic... (answered by MathLover1)
if the roots of the equation x^2+x+m=0 be real and unequal.Prove that the roots of the... (answered by Edwin McCravy)
If m and n are the roots of the equation {{{ ax^2+bx+c=0 }}} then what is the equation... (answered by richard1234)
If the Product of Two roots of quadratic equation x^2+mx +m(m-1)= 0 is 2,Then the value... (answered by Edwin McCravy)
Given that the quadratic equation x^2-2x-m+1=0 has 2 positive real number roots, α... (answered by ewatrrr)
for what value of m does the following equation has equal roots(2m+1)x^2 + 2(m+3)x +... (answered by josgarithmetic)
For what values of m has the equation x^2 - 2mx +8m -15 =0 a) one root? b) no roots?... (answered by MathLover1)
Given that {{{a}}} and {{{ma}}} are the roots of the equation {{{x^2+px+q=0}}}, show that (answered by KMST,MathTherapy)
Please solve using the roots: Given that the quadratic equation x^2-2x-m+1=0 has 2... (answered by Fombitz)