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^{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 = -aLet this be Equation I.

Hence, \alpha^{2} + \beta^{2} + \gamma^{2} = (\alpha + \beta + \gamma)^{2} - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 6Let 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









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