algebra « Mathematics Hothouse

Example 1:

Find the roots of $4x^{3}-16x^{2}-9x+36=0$ , given that one root is the negative of the another.

Solution 1:

If the roots are a, b, and c, we have $b=-a$ , say. So, using Viete’s relations,

$a-a+c=4$

$-a^{2}+ac-ac=-9/4$

$-a^{2}c = -9$ .

Hence, $c = 4$ and $a= 3/2 = -b$ . Hence, the roots are $\pm \frac{3}{2}, 4$ .

Example 2:

Let $a, b, c \in \Re$ , with $a \neq 0$ such that a and $4a+3b+2c$ have the same sign. Show that the equation $ax^{2}+bx+c=0$ cannot have both roots in the interval $(1,2)$ .

Solution 2:

Let $\alpha, \beta$ be roots of the given quadratic equation. We have $0 \leq \frac{4a+3b+2c}{a}$

which in turn equals

$= 4 + 3 \frac{b}{a} + 2\frac{c}{a} = 4 - 3(\alpha + \beta) + 2\alpha. \beta = (\alpha-1)(\beta-2) + (\alpha -2)(\beta -1)$ .

If $\alpha, \beta$ both belong to $(1,2)$ then each term of the sum will be negative, which is a contraction. Hence, the proof.

Example 3:

Consider all lines which meet the graph of $y = 2x^{4}+7x^{3}+3x-5$ in four distinct points, say $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), (x_{4}, y_{4})$ . Then, show that $\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}$ is independent of the line and find its value.

Solution 3:

Let $y=mx+c$ be any line which intersects the graph $y = 2x^{4}+7x^{3}+3x-5$ at $(x_{i}, y_{i})$ , where $1 \leq i \leq 4$ . Then, $x_{i}$ are the roots of

$mx + c = 2x^{4} + 7x^{3} + 3x -5$ .

(Note that $x_{i}'s$ are distinct as $x_{i} = x_{j}$ would imply $y_{i}=y_{j}$ ).

The above equation reduces to an equation of degree 4, namely,

$2x^{4}+7x^{3}+(3-m)x - 5-c=0$ . Hence, by Viete’s relations,

$\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} = - \frac{7}{8}$ .

Example 4:

The product of two of the four roots of $x^{4}-20x^{3}+kx^{2}+590x-1992=0$ is 24. Find k.

Solution 4:

Let the given equation be written as $f(x)=0$ , and let the roots of the equation be $r_{1}, r_{2}, r_{3}, r_{4}$ with $r_{1}r_{2}=24$ . Now,

$r_{1}r_{2}r_{3}r_{4}= -1992$ , so $r_{3}r_{4} = -1992/24 = -83$ . Also,

$f(x) = (x-r_{1})(x-r_{2})(x-r_{3})(x-r_{4}) = (x^{2}-cx+r_{1}r_{2})(x^{2}-dx+r_{3}r_{4}) = (x^{2}-cx+24)(x^{2}-dx-83)$

with $c = r_{1} + r_{2}$ , $d = r_{3} + r_{4}$ . Comparing coefficients of $x^{3}$ and x we get $c+d=20$ and $83c - 24d = 590$ . This gives $c = 10, d = 10$ . Comparing coefficients of $x^{2}$ , $k = cd - 83 + 24 = 100 -83 +24 =41$ .

Example 5:

If $\alpha$ and $\beta$ are roots of $x^{2}+px+q=0$ , where p and q are integers with $q|p^{2}$ , then show that

(i) $\alpha^{n} + \beta^{n}$ is an integer. ( $n \geq 1$ )

(ii) $\alpha^{n} + \beta^{n}$ is an integer divisible by q. ( $n \geq 2$ ).

Solution 5:

Since $\alpha, \beta$ are the roots of $x^{2}+px+q=0$ , we get

$\alpha + \beta = -p$ —– Equation A

$\alpha \beta = q$ —– Equation B

Note that $\alpha^{2} = -p \alpha - q$ . For $n \geq 2$ , multiplying this equation by $\alpha^{n-2}$ , we get $\alpha^{n} = -p \alpha^{n-1} - q\alpha^{n-2}$ . Similarly, $\beta^{n} = -p\beta^{n-1} -q\beta^{n-2}$ . Hence,

$\alpha^{n} + \beta^{n} = -p(\alpha^{n-1} + \beta^{n-1}) -q(\alpha^{n-2} +\beta^{n-2})$ Eqn C

Also, for $n=2$ , $\alpha^{2} + \beta^{2} = (-p)^{2} -2q = p^{2} -2q$ Equation D

1.By equation A and D, $\alpha + \beta$ and $\alpha^{2} + \beta^{2}$ are both integers. Hence, by Eqn C, it follows by induction on n that $\alpha^{n} + \beta^{n}$ is an integer for $n \geq 1$ .

2. Since $q|p^{2}$ , equation D shows that $q|\alpha^{2} + \beta^{2}$ . Further, $\alpha^{3} + \beta^{3} = -p(\alpha^{2} + \beta^{2}) - q(\alpha + \beta)$ and so $q|(\alpha^{3} + \beta^{3})$ . Hence, by C, it follows by induction on n that $q|(\alpha^{n}+\beta^{n})$ for $n \geq 2$ .

Example 6:

Find all integers a such that the equation $x^{3} - 3x +a=0$ has three integer roots.

Solution 6:

Let the integer roots of the given equation be $\alpha, \beta, \gamma$ . Then,

$\alpha + \beta + \gamma = 0, \alpha \beta + \beta \gamma + \gamma \alpha = -3$ , $\alpha \beta \gamma = -a$ . Let this be Equation I.

Hence, $\alpha^{2} + \beta^{2} + \gamma^{2} = (\alpha + \beta + \gamma)^{2} - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 6$ . Let this be Equation II.