Category Archives: algebra

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Non-linear equations for IITJEE Mathematics training

(Non-linear) equations involving three or more variables can only be solved in special cases. We shall here consider some of the most useful methods of solution.

Example 1.

Solve the following system of equations:

x+y+z=13 \ldots \ldots \ldots Equation I

x^{2}+y^{2}+z^{2}=65 \ldots \ldots \ldots Equation II

xy=10 \ldots \ldots\ldots Equation III

Solution I:

From II and III, we get (x+y)^{2} + z^{2}=85

Plug in u=x+y; then this equation becomes u^{2}+z^{2}=85.

Also, from I, we get u+z=13

hence, we obtain u=7 \hspace{0.1in}or \hspace{0.1in}u=6, z=6 \hspace{0.1in} or \hspace{0.1in}7.Thus,we have

x+y=7, xy=10; and x+y=6, xy=10. Hence, the solutions are

x=5, \hspace{0.1in} or \hspace{0.1in} 2; y=2, \hspace{0.1in} or \hspace{0.1in} 5; z=6;


x=3 \pm \sqrt{-1}; y=3 \mp \sqrt{-1}; z=7.

Example 2:

Solve the following system of equations:

(x+y)(x+z)=30 Equation I

(y+z)(y+x)=15 Equation II

(z+x)(z+y)=18 Equation III

Solution: 2:

Write u, v, w for y+z, z+x, x+y respectively; thus

vw=30, uw=15, vu=18… this equations A

Multiplying these equations together, we have

u^{2}v^{2}w^{2}=30 \times 15 \times 18 = 8100. Hence, uvw=\pm 90.

Combining this result with each of the equations in A, we have

u=3, v=6. w=5; or, u=-3, v=-6, w=-5; therefore, we get two sets of linear equations, as follows:




for which the solution set is: x=4, y=1, z=2; and the other set is:




for which the solution set if x=-4, y=-1, z=-2.

Example 3:

Solve the following system of equations:

y^{2}+yz+z^{2}=49…call this equation (1)

z^{2}+xz+x^{2}=19…call this equation (2)

x^{2}+xy+y^{2}=39…call this equation (3)

Subtracting (2) from (1), we get the following:

y^{2}-x^{2}+z(y-x)=30, that is, (y-x)(x+y+z)=30…call this equation (4).

Similarly, from (1) and (3), we get the following:

(z-x)(x+y+z)=10…call this equation (5).

Hence, from equations (4) and (5), by division, we get the following:

\frac{y-x}{z-x}=3, hence, y=3x-2z

Substituting in Equation (3), we obtain x^{2}-3xz+3z^{2}=13.

From (2), we get the following: x^{2}+xz+z^{2}=19

Solving these homogeneous equations (hint: put z=mx, where m is parameter to be found), we get the following:

x=\pm 2, z=\pm 3; and therefore, y=\pm 5

or, x=\pm \frac{11}{\sqrt{7}}; z=\pm \frac{1}{\sqrt{7}}; and therefore, y=\mp \frac{19}{\sqrt{7}}.

Example 4:

Solve the following system of equations:




Solution 4:

Multiply the equations by y, z, and x respectively, and add; then,

c^{2}x+a^{2}y+b^{2}z=0…call this equation I

Multiply the equations by z, x and y respectively; and add; then

b^{2}x+c^{2}y+a^{2}z=0…call this equation II

From (I) and (II), by cross multiplication,

\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{x}{c^{4}-a^{2}b^{2}}=k, suppose.

Substitute in any one of the given equations; then,

k^{2}(a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2})=1; hence, we get the following solution:

\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{z}{c^{4}-a^{2}b^{2}}=\pm \frac{1}{\sqrt{a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}}}.

More stuff in the pipeline,

Nalin Pithwa


Miscellaneous Example of Algebra

In order to find values of expressions involving a, b, c when these quantities are connected by the equation a+b+c=0, we might employ the substitution:

a = h + k

b = {\omega}h + {\omega^{2}}k

c = {\omega^{2}}h + {\omega}k

If, however, the expressions involve a, b, c symmetrically the method exhibited by the following example is preferable:


If a+b+c=0, show that 6(a^{5}+b^{5}+c^{5}) = 5(a^{3}+b^{3}+c^{3})(a^{2}+b^{2}+c^{2})


We have identically

(1+ax)(1+bx)(1+cx) = 1 + px + qx^{2}+rx^{3}, where p = a+b+c, q = ab + bc + ca, r = abc.

Hence, using the condition given, (1+ax)(1+bx)(1+cx) = 1+qx^{2}+rx^{3}

Taking logarithms and equating the coefficients of x^{n}, we have \frac{(-1)^{n-1}}{n} (a^{n}+b^{n}+c^{n}), which is equal to the coefficient of x^{n} in the expansion of \log {(1+qx^{2}+rx^{3})}, which is equal to the coefficient of x^{n} in the following:

(qx^{2} + rx^{3}) - \frac{1}{2}(qx^{2}+rx^{3})^{3} + \frac{1}{3}(qx^{2}+rx^{3})^{3} - \ldots

By putting n = 2, 3, 5 we obtain

q = -\frac{a^{2} + b^{2} + c^{2}}{2}

r = \frac{a^{3} + b^{3} + c^{3}}{3}

qr = - \frac{a^{5}+b^{5}+c^{5}}{5}, and hence, we get

 \frac{a^{5}+b^{5}+c^{5}}{5} = \frac{a^{3}+b^{3}+c^{3}}{3}  \times \frac{a^{2}+b^{2}+c^{2}}{2}, and the required result follows at once.

More stuff to follow…

Nalin Pithwa

Miscellaneous Examples of Algebra: Part 3 for IITJEE Mains

Many identities can be readily established by making use of  the properties of the cube roots of unity; as usual these will be denoted by 1, \omega, \omega^{2}.

Problem 1:

Show that (x+y)^{7} - x^{7} - y^{7} = 7xy(x+y)(x^{2}+xy+y^{2})^{2}

Solution 1:

The expression, E, on the left vanishes when x=0, y=0, x+y=0; hence, it must contain xy(x+y) as a factor.

Putting  x = {\omega}y, we have

E=((1+\omega)^{7}-\omega^{7}-1)y^{7} = ((-\omega^{2})^{7}-\omega^{7}-1)y^{7}=(-\omega^{2}-\omega-1)y^{7}

Hence, E contains x - {\omega}y as a factor; and similarly, we may show that it contains x - \omega^{2}y as a factor; that is, E is divisible by

(x-{\omega}y)(x-\omega^{2}y), or (x^{2}+xy+y^{2}).

Further, E being of seven, and xy(x+y)(x^{2}+xy+y^{2}) of five dimensions, the remaining factor must be of the form A(x^{2}+y^{2})+Bxy, thus,

(x+y)^{7}-x^{7}-y^{7} = xy(x+y)(x^{2}+xy+y^{2})(Ax^{2}+Bxy+Ay^{2}).

Putting x=1, y=1, we have 21 = 2A + B; putting x = 2, y = -1, we have 21= 5A-2B, and hence, A=7, B=7;

(x+y)^{7} - x^{7} - y^{7} = 7xy(x+y)(x^{2}+xy+y^{2}).

Problem 2:

Show that the product of a^{3} + b^{3} + c^{3} -3abc and x^{3} + y^{3} + z^{3} -3xyz can be put in the form A^{3} + B^{3} + C^{3} -3ABC.

Solution 2:

The product = (a+b+c)(a+{\omega}b + \omega^{2}c)(a + \omega^{2}b + {\omega}c) \times (x+y+z)(x+{\omega}y+\omega^{2}z)(x+\omega^{2}y + {\omega}z)

By taking these six factors in the pairs (a+b+c)(x+y+z),


and (a + \omega^{2}b + {\omega}c)(x + {\omega}y + \omega^{2}z),

we obtain the partial products:

A+B+C, A + {\omega}B + \omega^{2}C, and A + \omega^{2}B + {\omega}C

where  A =ax + by + cz, B = bx + cy + az, and C = cx + ay + bz

Thus, the product  = (A + B + C)(A + {\omega}B + \omega^{2}C)(A + \omega^{2}B + {\omega}C)

which, in turn, equals A^{3} + B^{3} + C^{3} + 3ABC

More esoteric algebraic miscellany is planned for you!

Nalin Pithwa

Miscellaneous examples of Algebra: part 2 for IITJEE Mains

Problem 1:

Prove that (a+b)^{5} - a^{5} - b^{5} = 5ab(a+b)(a^{2} + ab+b^{2}).

Proof I:

First carry out the following steps:

(a) Is the given expression symmetric?

(b) Is the given expression alternating?

(c) Is the given expression cyclic?

(d) Is the given expression homogeneous?

So here goes the proof:

Denote the expression by E. Clearly, the substitutions a=0, b=0 give us E=0. Hence, two of the factors are a, b. Similarly, the substitution a=-b give us E=0. So the other factor is (a+b).

The degree of  the given expression E is 5. E is symmetric w.r.t. a and b; it is homogeneous. The factors we found above are a, b, (a+b). So E=kab(a+b)F(x,y) where F(x,y) ought to be of degree 2, homogeneous and symmetric. So, let F(x, y) = Aa^{2}+Bab+Bb^{2}; thus, we have

(a+b)^{5}-a^{5}-b^{5}=kab(a+b)(Aa^{2}+Bab+Ab^{2}). Now, find k, A, and B by the undetermined coefficients.

Hence, you will get (a+b)^{5}-a^{5}-b^{5}=5ab(a+b)(a^{2}+ab+b^{2}).

Problem 2:

Find the factors of (b^{3}+c^{3})(b-c)+(c^{3}+a^{3})(c-a)+(a^{3}+b^{3})(a-b).

Solution 2:

First carry out the following steps:

(a) Is the given expression symmetric w.r.t. a and b; b and c; c and a?

(b) Is the given expression alternating w.r.t. a and b; b and c; c and a?

(c) Is the given expression cyclic w.r.t. a, b and c?

(d) Is the given expression homogeneous and if so, what is the degree of each expression?

Denote the expression by E; then, E is a function of a which vanishes when a=b, b=c, c=a, so the following are certainly factors of E: (a-b), (b-c), (c-a), and hence, E contains (a-b)(b-c)(c-a) as a factor.

Now note that E is of fourth degree and symmetric, so the remaining factor is of the form : M(a+b+c), where M is a coefficient to be determined.

So, we have now: E=M(a-b)(b-c)(c-a)(a+b+c), and you can easily find that M=1, and hence,

(b^{3}+c^{3})(b-c)+(c^{3}+a^{3})(c-a)+(a^{3}+b^{3})(a-b) = (a-b)(b-c)(c-a)(a+b+c)

More mathematical miscellany for IITJEE mathematics to follow!

Nalin Pithwa

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