Answer :
Let's address the questions one by one.
### Question 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.
First, expand the exprerssion [tex]\((x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\)[/tex]:
[tex]\[ (x-a)(x-b) = x^2 - (a+b)x + ab \][/tex]
[tex]\[ (x-b)(x-c) = x^2 - (b+c)x + bc \][/tex]
[tex]\[ (x-c)(x-a) = x^2 - (c+a)x + ac \][/tex]
Adding these expressions together:
[tex]\[ (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = x^2 - (a+b)x + ab + x^2 - (b+c)x + bc + x^2 - (c+a)x + ac \][/tex]
Combine like terms:
[tex]\[ 3x^2 - (a+b+b+c+c+a)x + (ab + bc + ac) = 0 \][/tex]
Simplify further:
[tex]\[ 3x^2 - 2(a+b+c)x + (ab + bc + ac) = 0 \][/tex]
To prove that the roots are real, we evaluate 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], and [tex]\(C=ab + bc + ac\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = B^2 - 4AC \][/tex]
Substitute 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]
Simplify the discriminant:
[tex]\[ \Delta = 4(a+b+c)^2 - 12(ab + bc + ac) \][/tex]
Rewrite [tex]\(4(a+b+c)^2 - 12(ab + bc + ac)\)[/tex]:
[tex]\[ \Delta = 4[(a+b+c)^2 - 3(ab + bc + ac)] \][/tex]
Since the expression inside the brackets is the square of the symmetric polynomial form of [tex]\(a, b,\)[/tex] and [tex]\(c\)[/tex], it is non-negative, ensuring:
[tex]\[ \Delta \ge 0 \][/tex]
Since [tex]\(\Delta \ge 0\)[/tex], the roots of the equation are real.
Prove that the roots are equal if [tex]\(a=b=c\)[/tex]:
Substitute [tex]\(a = b = c\)[/tex] in:
[tex]\[ 3x^2 - 2(a+b+c)x + (ab + bc + ac) = 0 \][/tex]
Since [tex]\(a = b = c\)[/tex]:
[tex]\[ a+b+c = 3a \][/tex]
[tex]\[ ab + bc + ac = 3a^2 \][/tex]
The quadratic equation becomes:
[tex]\[ 3x^2 - 2(3a)x + 3a^2 = 0 \][/tex]
Simplify:
[tex]\[ 3x^2 - 6ax + 3a^2 = 0 \][/tex]
Divide out the common factor of 3:
[tex]\[ x^2 - 2ax + a^2 = 0 \][/tex]
Factorize:
[tex]\[ (x-a)^2 = 0 \][/tex]
Hence, the roots are both [tex]\(x = a\)[/tex], confirming that the roots are equal.
### Question 9:
If [tex]\(a, b, c\)[/tex] are rational and [tex]\(a + b + c = 0\)[/tex], show that the roots of [tex]\((b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0\)[/tex] are rational.
First, simplify the coefficients since [tex]\(a + b + c = 0\)[/tex]:
Given:
[tex]\[ b + c = -a \][/tex]
[tex]\[ c + a = -b \][/tex]
[tex]\[ a + b = -c \][/tex]
Substitute these into the quadratic equation:
[tex]\[ (b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0 \][/tex]
Now replace:
[tex]\[ (b+c-a) \rightarrow (-a-a) = -2a \][/tex]
[tex]\[ (c+a-b) \rightarrow (-b-b) = -2b \][/tex]
[tex]\[ (a+b-c) \rightarrow (-c-c) = -2c \][/tex]
Thus, the equation becomes:
[tex]\[ -2a x^2 -2b x -2c = 0 \][/tex]
Divide the entire equation by -2 to simplify:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
Since [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are given as rational, and a quadratic equation with rational coefficients always has rational or complex conjugate roots, we need to confirm that the roots are indeed rational.
Given [tex]\(a + b + c = 0\)[/tex], the discriminant [tex]\(\Delta\)[/tex] of the quadratic equation is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
To ensure the roots are rational, we must show that [tex]\(\Delta\)[/tex] is a perfect square (since [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are rational).
Since [tex]\(a, b, c\)[/tex] are rational, [tex]\(b^2\)[/tex] and [tex]\(4ac\)[/tex] are rational, and if [tex]\(b^2 - 4ac\)[/tex] is non-negative and a perfect square, the roots [tex]\( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] will be rational.
In conclusion, both roots of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] are rational provided [tex]\(\Delta\)[/tex] is indeed a perfect square. Given [tex]\(a, b, c\)[/tex] are rational, hence the roots of the equation are rational.
### Question 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.
First, expand the exprerssion [tex]\((x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\)[/tex]:
[tex]\[ (x-a)(x-b) = x^2 - (a+b)x + ab \][/tex]
[tex]\[ (x-b)(x-c) = x^2 - (b+c)x + bc \][/tex]
[tex]\[ (x-c)(x-a) = x^2 - (c+a)x + ac \][/tex]
Adding these expressions together:
[tex]\[ (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = x^2 - (a+b)x + ab + x^2 - (b+c)x + bc + x^2 - (c+a)x + ac \][/tex]
Combine like terms:
[tex]\[ 3x^2 - (a+b+b+c+c+a)x + (ab + bc + ac) = 0 \][/tex]
Simplify further:
[tex]\[ 3x^2 - 2(a+b+c)x + (ab + bc + ac) = 0 \][/tex]
To prove that the roots are real, we evaluate 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], and [tex]\(C=ab + bc + ac\)[/tex].
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = B^2 - 4AC \][/tex]
Substitute 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]
Simplify the discriminant:
[tex]\[ \Delta = 4(a+b+c)^2 - 12(ab + bc + ac) \][/tex]
Rewrite [tex]\(4(a+b+c)^2 - 12(ab + bc + ac)\)[/tex]:
[tex]\[ \Delta = 4[(a+b+c)^2 - 3(ab + bc + ac)] \][/tex]
Since the expression inside the brackets is the square of the symmetric polynomial form of [tex]\(a, b,\)[/tex] and [tex]\(c\)[/tex], it is non-negative, ensuring:
[tex]\[ \Delta \ge 0 \][/tex]
Since [tex]\(\Delta \ge 0\)[/tex], the roots of the equation are real.
Prove that the roots are equal if [tex]\(a=b=c\)[/tex]:
Substitute [tex]\(a = b = c\)[/tex] in:
[tex]\[ 3x^2 - 2(a+b+c)x + (ab + bc + ac) = 0 \][/tex]
Since [tex]\(a = b = c\)[/tex]:
[tex]\[ a+b+c = 3a \][/tex]
[tex]\[ ab + bc + ac = 3a^2 \][/tex]
The quadratic equation becomes:
[tex]\[ 3x^2 - 2(3a)x + 3a^2 = 0 \][/tex]
Simplify:
[tex]\[ 3x^2 - 6ax + 3a^2 = 0 \][/tex]
Divide out the common factor of 3:
[tex]\[ x^2 - 2ax + a^2 = 0 \][/tex]
Factorize:
[tex]\[ (x-a)^2 = 0 \][/tex]
Hence, the roots are both [tex]\(x = a\)[/tex], confirming that the roots are equal.
### Question 9:
If [tex]\(a, b, c\)[/tex] are rational and [tex]\(a + b + c = 0\)[/tex], show that the roots of [tex]\((b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0\)[/tex] are rational.
First, simplify the coefficients since [tex]\(a + b + c = 0\)[/tex]:
Given:
[tex]\[ b + c = -a \][/tex]
[tex]\[ c + a = -b \][/tex]
[tex]\[ a + b = -c \][/tex]
Substitute these into the quadratic equation:
[tex]\[ (b+c-a)x^2 + (c+a-b)x + (a+b-c) = 0 \][/tex]
Now replace:
[tex]\[ (b+c-a) \rightarrow (-a-a) = -2a \][/tex]
[tex]\[ (c+a-b) \rightarrow (-b-b) = -2b \][/tex]
[tex]\[ (a+b-c) \rightarrow (-c-c) = -2c \][/tex]
Thus, the equation becomes:
[tex]\[ -2a x^2 -2b x -2c = 0 \][/tex]
Divide the entire equation by -2 to simplify:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
Since [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are given as rational, and a quadratic equation with rational coefficients always has rational or complex conjugate roots, we need to confirm that the roots are indeed rational.
Given [tex]\(a + b + c = 0\)[/tex], the discriminant [tex]\(\Delta\)[/tex] of the quadratic equation is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
To ensure the roots are rational, we must show that [tex]\(\Delta\)[/tex] is a perfect square (since [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are rational).
Since [tex]\(a, b, c\)[/tex] are rational, [tex]\(b^2\)[/tex] and [tex]\(4ac\)[/tex] are rational, and if [tex]\(b^2 - 4ac\)[/tex] is non-negative and a perfect square, the roots [tex]\( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] will be rational.
In conclusion, both roots of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] are rational provided [tex]\(\Delta\)[/tex] is indeed a perfect square. Given [tex]\(a, b, c\)[/tex] are rational, hence the roots of the equation are rational.
