10 Most FAQ’s of Theory of Equations Chapter in Inter 2nd Year Maths-2A (TS/AP)
7 Marks
LAQ-1 : Solve x4 – 10x3 + 26x2 – 10x + 1 = 0
The degree of the given equation is $$n=4$$ which is even. Also $$a_k = a_{n-k}$$ $$k = 0,1,2,3,4$$
So, the given equation is a standard reciprocal equation.
Now dividing the equation by x2 we get $$x^2 – 10x + 26 – \frac{10}{x} + \frac{1}{x^2} = 0$$
$$(x^2 + \frac{1}{x^2}) – 10(x + \frac{1}{x}) + 26 = 0$$
If $$x + \frac{1}{x} = y$$ then $$x^2 + \frac{1}{x^2} = y^2 – 2$$
$$[x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2 = y^2 – 2]$$
$$(y^2 – 2) – 10y + 26 = 0$$
$$y^2 – 10y + 24 = 0$$
$$y^2 – 6y – 4y + 24 = 0$$
$$y(y – 6) – 4(y – 6) = 0$$
$$(y – 4)(y – 6) = 0$$
$$y – 4 = 0$$ (or) $$y – 6 = 0$$
$$y = 4$$ (or) $$y = 6$$
If $$y = 4$$ then $$x + \frac{1}{x} = 4$$
$$x^2 + \frac{1}{x} = 4$$
$$x^2 + 1 = 4x$$
$$x^2 – 4x + 1 = 0$$
$$x = \frac{4 \pm \sqrt{(-4)^2 – 4(1)(1)}}{2 \cdot 1} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$$
If $$y = 6$$ then $$x + \frac{1}{x} = 6$$
$$x^2 + \frac{1}{x} = 6$$
$$x^2 + 1 = 6x$$
$$x^2 – 6x + 1 = 0$$
$$x = \frac{6 \pm \sqrt{(-6)^2 – 4(1)(1)}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$$
The roots of the given equation are $$2 \pm \sqrt{3}$$ $$3 \pm 2\sqrt{2}$$
LAQ-2 : Solve 2x4+x3-11x2+x+2=0
The degree of the given equation is $$n=4$$ which is even. Also, $$a_k = a_{n-k} \forall k = 0,1,2,3,4$$
So, the given equation is a standard reciprocal equation.
Now, dividing the equation by x2 we get $$2x^2 + x – 11 + \frac{1}{x} + \frac{2}{x^2} = 0$$
$$2\left(x^2 + \frac{1}{x^2}\right) + \left(x + \frac{1}{x}\right) – 11 = 0$$
If $$x + \frac{1}{x} = y$$ then $$x^2 + \frac{1}{x^2} = y^2 -2$$
$$2(y^2 – 2) + y – 11 = 0$$
$$2y^2 + y – 15 = 0$$
$$2y^2 + 6y – 5y – 15 = 0$$
$$2y(y + 3) – 5(y + 3) = 0$$
$$(2y – 5)(y + 3) = 0$$
$$2y – 5 = 0$$ (or) $$y + 3 = 0$$ $$y = \frac{5}{2}$$ $$y = -3$$
If $$y = \frac{5}{2}$$ then $$x + \frac{1}{x} = \frac{5}{2} = 2\frac{1}{2} = 2 + \frac{1}{2}$$
$$x = 2$$ or $$\frac{1}{2}$$
If $$y = -3$$ then $$x + \frac{1}{x} = -3$$
$$x^2 + \frac{1}{x} = -3$$
$$x^2 + 1 = -3x$$
$$x^2 + 3x + 1 = 0$$
$$x = -3 \pm \sqrt{3^2 – 4(1)(1)}/2(1) = -3 \pm \sqrt{9-4}/2 = -3 \pm \sqrt{5}/2$$
The roots of the given equation are $$2 \frac{1}{2} -3 + \sqrt{5}/2 -3 – \sqrt{5}/2$$
LAQ-3 : Solve 6x4-35x3+62x2-35x+6=0
The degree of the given equation is $$n=4$$ which is even. Also, $$a_k = a_{n-k} \, \forall \, k = 0,1,2,3,4$$
So, the given equation is a standard reciprocal equation.
Now, dividing the equation by x2 we get $$6x^2 – 35x + 62 – \frac{35}{x} + \frac{6}{x^2} = 0$$
$$\Rightarrow 6(x^2 + \frac{1}{x^2}) – 35(x + \frac{1}{x}) + 62 = 0$$
If $$x + \frac{1}{x} = y$$ then $$x^2 + \frac{1}{x^2} = y^2 – 2$$
$$\Rightarrow 6(y^2 – 2) – 35y + 62 = 0 \Rightarrow 6y^2 – 35y + 50 = 0 \Rightarrow 6y^2 – 20y – 15y + 50 = 0$$
$$\Rightarrow 2y(3y – 10) – 5(3y – 10) = 0 \Rightarrow (3y – 10)(2y – 5) = 0$$
$$\Rightarrow 3y – 10$$ (or) $$2y – 5 = 0 \Rightarrow y = \frac{10}{3} y = \frac{5}{2}$$
If $$y = \frac{10}{3}$$ then $$x + \frac{1}{x} = \frac{10}{3} = 3\frac{1}{3} = 3 + \frac{1}{3} \Rightarrow x = 3 \frac{1}{3}$$
If $$y = \frac{5}{2}$$ then $$x + \frac{1}{x} = \frac{5}{2} = 2\frac{1}{2} = 2 + \frac{1}{2} \Rightarrow x = 2 or \frac{1}{2}$$
The roots are $$3, \frac{1}{3}, 2, \frac{1}{2}$$
LAQ-4 : Solve 2x5+x4-12x3-12x2+x+2=0
The degree of the given equation is n = 5, which is odd. Also, $$a_k = a_{n-k} \, \forall \, k = 0,1,2,3,4,5$$
Hence, the given equation is a reciprocal equation of class 1 of odd degree.
-1 is a root of the given equation.
Now, dividing the expression by (x+1) we have
Now, we solve the SRE $$2x^4 – x^3 – 11x^2 – x + 2 = 0$$
On dividing the equation by x2 we get
$$2x^2 – x – 11 – \frac{1}{x} + \frac{2}{x^2} = 0$$
$$\Rightarrow 2(x^2 + \frac{1}{x^2}) – (x + \frac{1}{x}) – 11 = 0$$
Put $$x + \frac{1}{x} = y$$
$$\Rightarrow x^2 + \frac{1}{x^2} = y^2 – 2$$
$$\Rightarrow 2(y^2 – 2) – y – 11 = 0 \Rightarrow 2y^2 – 4 – y – 11 = 0$$
$$\Rightarrow 2y^2 – y – 15 = 0 \Rightarrow 2y^2 – 6y + 5y – 15 = 0$$
$$\Rightarrow 2y(y – 3) + 5(y – 3) = 0 \Rightarrow (y – 3)(2y + 5) = 0$$
$$\Rightarrow y – 3 = 0$$ (or) $$2y + 5 = 0 \Rightarrow y = 3 (or) y = -\frac{5}{2}$$
If $$y = 3$$ then $$x + \frac{1}{x} = 3 \Rightarrow x^2 + 1 = 3x \Rightarrow x^2 – 3x + 1 = 0$$
$$\Rightarrow x = \frac{3 \pm \sqrt{(-3)^2 – 4\cdot1\cdot1}}{2\cdot1} = \frac{3 \pm \sqrt{9 – 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$
If $$y = -\frac{5}{2}$$ then $$x + \frac{1}{x} = -\frac{5}{2} \Rightarrow x^2 + 1 = -\frac{5x}{2} \Rightarrow 2x^2 + 2 = -5x$$
$$x = -2 or x = -\frac{1}{2}$$
The roots of the given equation are $$-1 -2 -\frac{1}{2} \frac{3 \pm \sqrt{5}}{2}$$
LAQ-5 : Solve 6x6-25x5+31x4-31x2+25x-6=0
The degree of the given equation is n = 6 which is even. Also, $$a_k = -a_{n-k} \, \forall \, k = 0,1,2,3,4,5,6$$
Hence the given equation is a reciprocal equation of class II of even degree.
Hence 1,−1 are the roots of $$6x^6 -25x^5 + 31x^4 – 31x^2 + 25x – 6 = 0$$
On dividing the expression by $$(x-1) (x+1)$$ we have
Now, we solve the S.R.E $$6x^4 – 25x^3 + 37x^2 – 25x + 6 = 0$$
On dividing this equation by x2 we get
$$6x^2 – 25x + 37 – \frac{25}{x} + \frac{6}{x^2} = 0 \Rightarrow 6(x^2 + \frac{1}{x^2}) – 25(x + \frac{1}{x}) + 37 = 0$$
Put $$x + \frac{1}{x} = y$$ so that $$x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2 = y^2 – 2$$
$$\Rightarrow 6(y^2 – 2) – 25(y) + 37 = 0 \Rightarrow 6y^2 – 12 – 25y + 37 = 0$$
$$\Rightarrow 6y^2 – 25y + 25 = 0 \Rightarrow 6y^2 – 15y -10y + 25 = 0 \Rightarrow 3y(2y – 5) – 5(2y – 5) = 0$$
$$\Rightarrow (2y – 5)(3y – 5) = 0 \Rightarrow y = \frac{5}{2}$$
If $$y = \frac{5}{2}$$ then $$x + \frac{1}{x} = \frac{5}{2} = 2\frac{1}{2} = 2 + \frac{1}{2} \Rightarrow x = 2 (or) \frac{1}{2}$$
If $$y = \frac{5}{3}$$ then $$x + \frac{1}{x} = \frac{5}{3} \Rightarrow x^2 + \frac{1}{x} = \frac{5}{3}$$
$$\Rightarrow 3x^2 + 3 = 5x \Rightarrow 3x^2 – 5x + 3 = 0$$
$$\Rightarrow x = \frac{5 \pm \sqrt{(-5)^2 – 4(3)(3)}}{6} = \frac{5 \pm \sqrt{25 – 36}}{6} = \frac{5 \pm \sqrt{-11}}{6}$$
$$x = \frac{5 \pm \sqrt{11}i}{6}$$
Hence all the six roots of the given equation are $$1 -1 2 \frac{1}{2} \frac{5 \pm \sqrt{11}i}{6}$$
LAQ-6 : Solve the equation x5-5x4+9x3-9x2+5x-1=0
The degree of the given equation is n = 5, which is odd. Also, $$a_k = -a_{n-k} \, \forall \, k = 0,1,2,3,4,5$$
Hence the given equation is a reciprocal equation of class II of odd degree.
1 is a root of $$x^5 – 5x^4 + 9x^3 – 9x^2 + 5x – 1 = 0$$
On dividing the expression by (x−1) we have
Now, we solve the S.R.E $$x^4 – 4x^3 + 5x^2 – 4x + 1 = 0$$
On dividing this equation by x2 we get
$$x^2 – 4x + 5 – \frac{4}{x} + \frac{1}{x^2} = 0 \Rightarrow (x^2 + \frac{1}{x^2}) – 4(x + \frac{1}{x}) + 5 = 0$$
Put $$x + \frac{1}{x} = y$$ so that $$x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2 = y^2 – 2$$
$$\Rightarrow (y^2 – 2) – 4y + 5 = 0 \Rightarrow y^2 – 4y + 3 = 0$$
$$y^2 – 4y + 3 = 0 \Rightarrow (y – 3)(y – 1) = 0 \Rightarrow y = 3 or 1$$
If $$y = 3$$ then $$x + \frac{1}{x} = 3 \Rightarrow x^2 + \frac{1}{x} = 3 \Rightarrow x^2 + 1 = 3x \Rightarrow x^2 – 3x + 1 = 0$$
$$\Rightarrow x = \frac{3 \pm \sqrt{(-3)^2 – 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9 – 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$
If $$y = 1$$ then $$x + \frac{1}{x} = 1 \Rightarrow x^2 + \frac{1}{x} = 1 \Rightarrow x^2 + 1 = x \Rightarrow x^2 – x + 1 = 0$$
$$\Rightarrow x = \frac{1 \pm \sqrt{(-1)^2 – 4(1)(1)}}{2(1)} = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$$
Hence all the five roots of the given equation are $$1 3 \pm \sqrt{5}/2 and 1 \pm i\sqrt{3}/2$$
LAQ-7 : Solve the equation 8x3-36x2-18x+81=0 the roots being in A.P
Let the roots of $$8x^3 – 36x^2 – 18x + 81 = 0$$ in A.P be taken as $$a-d a a+d$$
Now $$S_1 = (a – d) + a + (a + d) = \frac{36}{8} = \frac{9}{2}$$
$$\Rightarrow 3a = \frac{9}{2} \Rightarrow a = \frac{3}{2}$$
$$S_3 = (a – d)a(a + d) = -\frac{81}{8}$$
$$\Rightarrow a(a^2 – d^2) = -\frac{81}{8}$$
$$\Rightarrow \frac{3}{2}(\frac{9}{4} – d^2) = -\frac{81}{8} \Rightarrow (\frac{9}{4} – d^2) = -\frac{81}{8} \times \frac{2}{3} = -\frac{27}{4}$$
$$\Rightarrow \frac{9}{4} – d^2 = -\frac{27}{4} \Rightarrow d^2 = \frac{9}{4} + \frac{27}{4} = \frac{36}{4} = 9$$
$$\Rightarrow d = \pm 3$$
The roots of the equation are $$a – d a a + d \Rightarrow \frac{3}{2} – 3 \frac{3}{2} \frac{3}{2} + 3$$
$$\Rightarrow -\frac{3}{2} \frac{3}{2} \frac{9}{2}$$
LAQ-8 : Solve 3x3-26x2+52x-24=0, given that the roots are in G.P
Let the roots of $$3x^3 – 26x^2 + 52x – 24 = 0$$ in GP be taken as $$a/r a ar$$
Product of roots $$S_3 = (a/r)(a)(ar) = 24/3 = 8 \Rightarrow a^3 = 8 \Rightarrow a = 2$$
$$S_1 = a/r + a + ar = 26/3 \Rightarrow 2(1/r + 1 + r) = 26/3 \Rightarrow 2(1 + r + r^2/r) = 26/3$$
$$\Rightarrow (1 + r + r^2/r) = 13/3 \Rightarrow 3(r^2 + r + 1) = 13r \Rightarrow 3r^2 + 3r + 3 = 13r \Rightarrow 3r^2 – 10r + 3 = 0$$
$$\Rightarrow 3r^2 – 9r – r + 3 = 0 \Rightarrow 3r(r-3) – 1(r-3) = 0 \Rightarrow (3r-1)(r-3) = 0 \Rightarrow r = 3 or 1/3$$
The roots are $$(a/r) (a) (ar)$$
$$\Rightarrow 2/3 2 2(3)$$
$$\Rightarrow 2/3 2 6$$
LAQ-9 : Solve 18x3+81x2+121x+60=0, given that a root is equal half the sum of the remaining roots
Let $$\alpha \beta \gamma$$ be the roots of $$18x^3 + 81x^2 + 121x + 60 = 0$$
Given that one root is half the sum of the remaining two roots. Let $$\beta = \frac{\alpha + \gamma}{2}$$
$$\Rightarrow \alpha \beta \gamma$$ are in A.P.
We take $$\alpha = a-d \beta = a \gamma = a+d$$
The given equation is $$18x^3 + 81x^2 + 121x + 60 = 0$$
$$S_1 = \alpha + \beta + \gamma = -\frac{b}{a} \Rightarrow (a – d) + a + (a + d) = -\frac{81}{18} = -\frac{9}{2} \Rightarrow 3a = -\frac{9}{2} \Rightarrow a = -\frac{3}{2}$$
$$S_3 = \alpha\beta\gamma = -\frac{d}{a} \Rightarrow (a – d)(a)(a + d) = -\frac{60}{18} = -\frac{10}{3} \Rightarrow a(a^2 – d^2) = -\frac{10}{3}$$
$$\Rightarrow -\frac{3}{2}\left(\left(-\frac{3}{2}\right)^2 – d^2\right) = -\frac{10}{3} \Rightarrow \frac{9}{4} – d^2 = \frac{10}{3} \times \frac{2}{3} = \frac{20}{9} \Rightarrow d^2 = \frac{9}{4} – \frac{20}{9} = \frac{81 – 80}{36} = \frac{1}{36} \Rightarrow d = \frac{1}{6}$$
Now $$a – d = -\frac{3}{2} – \frac{1}{6} = -\frac{9}{6} – \frac{1}{6} = -\frac{10}{6} = -\frac{5}{3}$$
Also $$a + d = -\frac{3}{2} + \frac{1}{6} = -\frac{9}{6} + \frac{1}{6} = -\frac{8}{6} = -\frac{4}{3}$$
The roots $$a-d a a+d$$ are $$-\frac{5}{3} -\frac{3}{2} -\frac{4}{3}$$
LAQ-10 : Solve the equation x4-4x2+8x+35=0, given that 2 + i√3 is a root of the equation
Let $$f(x) = x^4 – 4x^2 + 8x + 35$$
$$2 + i\sqrt{3}$$ is a root of $$f(x) = 0 \Rightarrow 2 – i\sqrt{3}$$ is also a root of $$f(x) = 0$$
The sum of roots $$(2 + i\sqrt{3}) + (2 – i\sqrt{3}) = 4$$ and product of roots $$(2 + i\sqrt{3})(2 – i\sqrt{3}) = 4 + 3 = 7$$
The equation with roots $$2 + \sqrt{3} 2 – \sqrt{3}$$ is $$x^2 – (\text{sum of roots})x + \text{product} = 0$$
$$\Rightarrow x^2 – 4x + 7 = 0$$
$$\Rightarrow x^2 – 4x + 7$$ is a factor of $$f(x)$$
Now, $$x^2 + 4x + 5 = 0 \Rightarrow x = -4 \pm \sqrt{16 – 20}/2 = -4 \pm \sqrt{-4}/2 = -2 \pm i$$
The roots of the given equation are $$2 + i\sqrt{3} 2 – i\sqrt{3} -2 + i -2 -i$$