Answer :
Let's tackle each part of the problem step by step.
### 8. Prove that the roots of the equation [tex]\((x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\)[/tex] are real. Also, prove that the roots are equal if [tex]\(a=b=c\)[/tex].
Let's denote the given equation as [tex]\(f(x) = (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)\)[/tex].
First, expand each term:
1. [tex]\((x-a)(x-b) = x^2 - x(b + a) + ab\)[/tex]
2. [tex]\((x-b)(x-c) = x^2 - x(c + b) + bc\)[/tex]
3. [tex]\((x-c)(x-a) = x^2 - x(a + c) + ac\)[/tex]
Summing these three expressions, we get:
[tex]\[f(x) = [x^2 - x(b + a) + ab] + [x^2 - x(c + b) + bc] + [x^2 - x(a + c) + ac]\][/tex]
Combining like terms, we obtain:
[tex]\[ f(x) = 3x^2 - x[ (b + a) + (c + b) + (a + c) ] + [ ab + bc + ac ] \][/tex]
[tex]\[ f(x) = 3x^2 - x[ 2(a + b + c) ] + [ ab + bc + ac ] \][/tex]
[tex]\[ f(x) = 3x^2 - 2x(a + b + c) + (ab + bc + ac) \][/tex]
To simplify the analysis of the roots, consider the case [tex]\( a = b = c \)[/tex].
### Case 1: [tex]\( a = b = c \)[/tex]
Plug [tex]\(a = b = c\)[/tex] into the function [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = 3x^2 - 2x(a + a + a) + (a^2 + a^2 + a^2) \][/tex]
[tex]\[ f(x) = 3x^2 - 6ax + 3a^2 \][/tex]
Divide through by 3:
[tex]\[ f(x) = x^2 - 2ax + a^2 \][/tex]
This simplifies to:
[tex]\[ f(x) = (x-a)^2 \][/tex]
The roots of this function are where the function equals zero, so:
[tex]\[ (x-a)^2 = 0 \][/tex]
The solution to this equation is:
[tex]\[ x = a \][/tex]
This root is real and equal, proving that if [tex]\(a = b = c\)[/tex], the roots are real and equal.
### Case 2: General [tex]\( a, b, c \)[/tex]
Now let's investigate the nature of the roots for general values of [tex]\(a, b, c\)[/tex]. We will solve:
[tex]\[ 3x^2 - 2x(a + b + c) + (ab + bc + ac) = 0 \][/tex]
To determine if the roots are real, we use the discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex], where:
[tex]\[ A = 3 \][/tex]
[tex]\[ B = -2(a + b + c) \][/tex]
[tex]\[ C = ab + bc + ac \][/tex]
The discriminant of a quadratic equation is given by:
[tex]\[ \Delta = B^2 - 4AC \][/tex]
Substituting the values of [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
[tex]\[ \Delta = [-2(a + b + c)]^2 - 4 \cdot 3 \cdot (ab + bc + ac) \][/tex]
[tex]\[ \Delta = 4(a + b + c)^2 - 12(ab + bc + ac) \][/tex]
[tex]\[ \Delta = 4[(a + b + c)^2 - 3(ab + bc + ac)] \][/tex]
Thus, the roots of the equation will be real if the discriminant is non-negative:
[tex]\[ \Delta = 4[(a + b + c)^2 - 3(ab + bc + ac)] \geq 0 \][/tex]
Since the discriminant [tex]\(\Delta\)[/tex] is expressed as a multiple of 4 but for the roots to be real, it is required that [tex]\((a + b + c)^2 - 3(ab + bc + ac)\)[/tex] be greater than or equal to zero.
### 9. Show that the roots of the equation [tex]\((b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0\)[/tex] are rational if [tex]\(a + b + c = 0\)[/tex] and [tex]\(a, b, c\)[/tex] are rational.
Given:
[tex]\[ (b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0 \][/tex]
Since [tex]\(a, b, c\)[/tex] are rational numbers, let's substitute [tex]\( a + b + c = 0 \)[/tex].
From [tex]\( a + b + c = 0 \)[/tex], we get [tex]\( c = - (a + b) \)[/tex].
Substitute [tex]\( c = - (a + b) \)[/tex] into the coefficients of the quadratic equation:
- For the coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ b + c - a = b - (a + b) - a = -2a \][/tex]
- For the coefficient of [tex]\(x\)[/tex]:
[tex]\[ c + a - b = - (a + b) + a - b = - 2b \][/tex]
- For the constant term:
[tex]\[ a + b - c = a + b - (-(a + b)) = a + b + a + b = 2(a + b) \][/tex]
Thus, the equation reduces to:
[tex]\[ -2ax^2 - 2bx + 2(a + b) = 0 \][/tex]
Divide the equation by 2:
[tex]\[ -ax^2 - bx + (a + b) = 0 \][/tex]
Now, let's change all signs to make it more simplified:
[tex]\[ ax^2 + bx - (a + b) = 0 \][/tex]
We can use the quadratic formula [tex]\(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)[/tex], where [tex]\(A = a\)[/tex], [tex]\(B = b\)[/tex], and [tex]\(C = -(a + b)\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = b^2 - 4a(- (a + b)) \][/tex]
[tex]\[ \Delta = b^2 + 4a(a + b) \][/tex]
[tex]\[ \Delta = b^2 + 4a^2 + 4ab \][/tex]
[tex]\[ \Delta = (2a + b)^2 \][/tex]
Since [tex]\(\Delta = (2a + b)^2\)[/tex] is a perfect square, the roots are rational.
Thus, if [tex]\(a, b, c\)[/tex] are rational and [tex]\(a + b + c = 0\)[/tex], the roots of the equation [tex]\((b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0\)[/tex] are rational.
### 8. Prove that the roots of the equation [tex]\((x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\)[/tex] are real. Also, prove that the roots are equal if [tex]\(a=b=c\)[/tex].
Let's denote the given equation as [tex]\(f(x) = (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)\)[/tex].
First, expand each term:
1. [tex]\((x-a)(x-b) = x^2 - x(b + a) + ab\)[/tex]
2. [tex]\((x-b)(x-c) = x^2 - x(c + b) + bc\)[/tex]
3. [tex]\((x-c)(x-a) = x^2 - x(a + c) + ac\)[/tex]
Summing these three expressions, we get:
[tex]\[f(x) = [x^2 - x(b + a) + ab] + [x^2 - x(c + b) + bc] + [x^2 - x(a + c) + ac]\][/tex]
Combining like terms, we obtain:
[tex]\[ f(x) = 3x^2 - x[ (b + a) + (c + b) + (a + c) ] + [ ab + bc + ac ] \][/tex]
[tex]\[ f(x) = 3x^2 - x[ 2(a + b + c) ] + [ ab + bc + ac ] \][/tex]
[tex]\[ f(x) = 3x^2 - 2x(a + b + c) + (ab + bc + ac) \][/tex]
To simplify the analysis of the roots, consider the case [tex]\( a = b = c \)[/tex].
### Case 1: [tex]\( a = b = c \)[/tex]
Plug [tex]\(a = b = c\)[/tex] into the function [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = 3x^2 - 2x(a + a + a) + (a^2 + a^2 + a^2) \][/tex]
[tex]\[ f(x) = 3x^2 - 6ax + 3a^2 \][/tex]
Divide through by 3:
[tex]\[ f(x) = x^2 - 2ax + a^2 \][/tex]
This simplifies to:
[tex]\[ f(x) = (x-a)^2 \][/tex]
The roots of this function are where the function equals zero, so:
[tex]\[ (x-a)^2 = 0 \][/tex]
The solution to this equation is:
[tex]\[ x = a \][/tex]
This root is real and equal, proving that if [tex]\(a = b = c\)[/tex], the roots are real and equal.
### Case 2: General [tex]\( a, b, c \)[/tex]
Now let's investigate the nature of the roots for general values of [tex]\(a, b, c\)[/tex]. We will solve:
[tex]\[ 3x^2 - 2x(a + b + c) + (ab + bc + ac) = 0 \][/tex]
To determine if the roots are real, we use the discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex], where:
[tex]\[ A = 3 \][/tex]
[tex]\[ B = -2(a + b + c) \][/tex]
[tex]\[ C = ab + bc + ac \][/tex]
The discriminant of a quadratic equation is given by:
[tex]\[ \Delta = B^2 - 4AC \][/tex]
Substituting the values of [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
[tex]\[ \Delta = [-2(a + b + c)]^2 - 4 \cdot 3 \cdot (ab + bc + ac) \][/tex]
[tex]\[ \Delta = 4(a + b + c)^2 - 12(ab + bc + ac) \][/tex]
[tex]\[ \Delta = 4[(a + b + c)^2 - 3(ab + bc + ac)] \][/tex]
Thus, the roots of the equation will be real if the discriminant is non-negative:
[tex]\[ \Delta = 4[(a + b + c)^2 - 3(ab + bc + ac)] \geq 0 \][/tex]
Since the discriminant [tex]\(\Delta\)[/tex] is expressed as a multiple of 4 but for the roots to be real, it is required that [tex]\((a + b + c)^2 - 3(ab + bc + ac)\)[/tex] be greater than or equal to zero.
### 9. Show that the roots of the equation [tex]\((b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0\)[/tex] are rational if [tex]\(a + b + c = 0\)[/tex] and [tex]\(a, b, c\)[/tex] are rational.
Given:
[tex]\[ (b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0 \][/tex]
Since [tex]\(a, b, c\)[/tex] are rational numbers, let's substitute [tex]\( a + b + c = 0 \)[/tex].
From [tex]\( a + b + c = 0 \)[/tex], we get [tex]\( c = - (a + b) \)[/tex].
Substitute [tex]\( c = - (a + b) \)[/tex] into the coefficients of the quadratic equation:
- For the coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ b + c - a = b - (a + b) - a = -2a \][/tex]
- For the coefficient of [tex]\(x\)[/tex]:
[tex]\[ c + a - b = - (a + b) + a - b = - 2b \][/tex]
- For the constant term:
[tex]\[ a + b - c = a + b - (-(a + b)) = a + b + a + b = 2(a + b) \][/tex]
Thus, the equation reduces to:
[tex]\[ -2ax^2 - 2bx + 2(a + b) = 0 \][/tex]
Divide the equation by 2:
[tex]\[ -ax^2 - bx + (a + b) = 0 \][/tex]
Now, let's change all signs to make it more simplified:
[tex]\[ ax^2 + bx - (a + b) = 0 \][/tex]
We can use the quadratic formula [tex]\(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)[/tex], where [tex]\(A = a\)[/tex], [tex]\(B = b\)[/tex], and [tex]\(C = -(a + b)\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = b^2 - 4a(- (a + b)) \][/tex]
[tex]\[ \Delta = b^2 + 4a(a + b) \][/tex]
[tex]\[ \Delta = b^2 + 4a^2 + 4ab \][/tex]
[tex]\[ \Delta = (2a + b)^2 \][/tex]
Since [tex]\(\Delta = (2a + b)^2\)[/tex] is a perfect square, the roots are rational.
Thus, if [tex]\(a, b, c\)[/tex] are rational and [tex]\(a + b + c = 0\)[/tex], the roots of the equation [tex]\((b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0\)[/tex] are rational.
