## Tag Archives: algebra

### Miscellaneous examples of Algebra Part I: IIT JEE Mains

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.

So, $0 \leq \alpha^{2}, \beta^{2}, \gamma^{2} \leq 6$ and so the solution of Equation II are essentially the following:

$\alpha = 2, \beta = -1, \gamma = -1$ —- Call this ***

$\alpha = -2, \beta =1, \gamma = 1$ —- Call this *****

Both these sets satisfy Equation I. Hence, the required values of a are $a = -2, 2$ corresponding to the roots in (***) and (*****) respectively.

Example 7:

Find the remainder when $(x+1)^{n}$ is divided by $(x-1)^{3}$.

Solution 7:

Dividing $(x+1)^{n}$ by $(x-1)^{3}$, we get $(x+1)^{n} = f(x)(x-1)^{3}$, we get

$(x+1)^{n} = f(x) (x-1)^{3} + Ax^{2} + Bx + C$.

Put $x-1=y$ or $x=y+1$. Hence,

$(y+2)^{n} = f(y+1)y^{3} + A(y+1)^{2} + B(y+1) + C$

Using Binomial theorem, we get

$y^{n} + \ldots + y^{2}(\frac{n(n-1)}{2}2^{n-2}) + y(n2^{n-1}) + 2^{n}$, which in turn equals

$= f(y+1)y^{3}+ Ay^{2} + (2A + B)y + A + B + C$ Call this Equation @@@

Now, equating coefficients of $y^{2}, y^{1}, y^{0}$ we get

$A = n(n-1)2^{n-3}$, $2A + B = n.2^{n-1}$, $A+B+C=2^{n}$.

Solving these equations, we get

$A = n(n-1)2^{n-3}$, $B = n(3-n)2^{n-2}$, $C = (n^{2}-5n+8)2^{n-3}$

Hence, the remainder is

$n(n-1)2^{n-3}x^{2} + n(3-n)2^{n-2}x + (n^{2}-5n+8)2^{n-3}$

More algebraic stuff in the pipeline!!

Nalin Pithwa