11 Most VSAQ’s of Theory of Equations Chapter in Inter 2nd Year Maths-2A (TS/AP)

2 Marks

VSAQ-1 : If 1,1,α are the roots of x³-6x²+9x-4=0 then find α

From the given equation we get $$a_0 = 1$$ $$a_1 = -6$$ $$a_3 = -4$$

Product of roots $$1\cdot1\cdot\alpha = S_3 = -\frac{a_3}{a_0} = \frac{4}{1}$$

$$\alpha = 4$$

---

VSAQ-2 : If -1,2,α are the roots of 2x³+x²-7x-6=0 then find α

From the given equation we get $$a_0 = 2$$ $$a_1 = 1$$ $$a_2 = -7$$ $$a_3 = -6$$

Sum of the roots

$$S_1 = -1 + 2 + \alpha = \frac{a_1}{a_0} = -\frac{1}{2}$$

$$\Rightarrow 1 + \alpha = -\frac{1}{2}$$

$$\Rightarrow \alpha = -1 – \frac{1}{2} = -\frac{3}{2}$$

---

VSAQ-3 : If 1,-2,3 are the roots of x³-2x²+ax+6=0 then find a

1 is a root of $$x^3 – 2x^2 + ax + 6 = 0$$

$$\Rightarrow 1^3 – 2(1^2) + a(1) + 6 = 0$$

$$\Rightarrow 1 – 2 + a + 6 = 0$$

$$\Rightarrow a + 5 = 0$$

$$\Rightarrow a = -5$$

---

VSAQ-4 : If the product of the roots of 4x³+16x²-9x-a=0 is 9, then find a

From the given equation, we get $$a_0 = 4$$ $$a_1 = 16$$ $$a_2 = -9$$ $$a_3 = -a$$

The product of the roots is 9.

$$\Rightarrow S_3 = -\frac{a_3}{a_0} = 9$$

$$\Rightarrow \frac{a}{4} = 9$$

$$\Rightarrow a = 4 \times 9 = 36$$

---

VSAQ-5 : If α,β,1 are the roots of x³-2x²+5x+6=0 then find α,β

From the given equation, we get $$a_0 = 1$$ $$a_1 = -2$$ $$a_2 = 5$$ $$a_3 = 6$$

Sum of the roots $$S_1 = \alpha + \beta + 1 = -\frac{a_1}{a_0} = 2/1 = 2$$

$$\Rightarrow \alpha + \beta = 2 – 1 = 1$$

Product of the roots $$S_3 = \alpha\cdot\beta\cdot1 = -\frac{a_3}{a_0} = -6/1 = -6$$

$$\Rightarrow \alpha\beta = -6$$

$$\alpha + \beta = 1 \alpha\beta = -6$$

Solving the above two equations, we get $$\alpha=3 \beta=-2$$

---

VSAQ-6 : Find polynomial equation whose roots are reciprocals of roots of x⁴-3x³+7x²+5x-2=0

Let $$f(x) = x^4 – 3x^3 + 7x^2 + 5x – 2 = 0$$

Reciprocal equation is $$f\left(\frac{1}{x}\right) = 0$$

$$\Rightarrow \frac{1}{x^4} – \frac{3}{x^3} + \frac{7}{x^2} + \frac{5}{x} – 2 = 0$$

$$\Rightarrow 1 – 3x + 7x^2 + 5x^3 – 2x^4 = 0$$

$$\Rightarrow 2x^4 – 5x^3 – 7x^2 + 3x – 1 = 0$$

---

VSAQ-7 : Find the algebraic equation whose roots are two times the roots of x⁵-2x⁴+3x³-2x²+4x+3=0

Let $$f(x) = x^5 – 2x^4 + 3x^3 – 2x^2 + 4x + 3$$

The required equation for $$f\left(\frac{x}{2}\right) = 0$$

$$\Rightarrow \left(\frac{x}{2}\right)^5 – 2\left(\frac{x}{2}\right)^4 + 3\left(\frac{x}{2}\right)^3 – 2\left(\frac{x}{2}\right)^2 + 4\left(\frac{x}{2}\right) + 3 = 0$$

$$\Rightarrow \frac{1}{32}x^5 – \frac{2}{16}x^4 + \frac{3}{8}x^3 – \frac{2}{4}x^2 + \frac{4}{2}x + 3 = 0$$

$$\Rightarrow x^5 – 4x^4 + 12x^3 – 16x^2 + 64x + 96 = 0$$

---

VSAQ-8 : Find an algebraic equation of degree 4 whose roots are 3 times the roots of the equation 6x⁴-7x³+8x²-7x+2=0

$$f\left(\frac{x}{3}\right) = 0$$

$$\Rightarrow 6\left(\frac{x}{3}\right)^4 – 7\left(\frac{x}{3}\right)^3 + 8\left(\frac{x}{3}\right)^2 – 7\left(\frac{x}{3}\right) + 2 = 0$$

$$\Rightarrow \frac{1}{81}[6x^4 – 21x^3 + 72x^2 – 189x + 162] = 0$$

$$\Rightarrow 6x^4 – 21x^3 + 72x^2 – 189x + 162 = 0$$

---

VSAQ-9 : Find transformed equation whose roots are negatives of the roots of x⁴+5x³+11x+3=0

Given the function $$f(x) = x^4 + 5x^3 + 11x + 3$$

$$f(-x) = (-x)^4 + 5(-x)^3 + 11(-x) + 3 = 0$$

$$x^4 – 5x^3 – 11x + 3 = 0$$

---

VSAQ-10 : Find the monomic polynomial equation of the degree 3 whose roots are 2,3 and 6

Given the requirement to express the monic polynomial equation $$(x – 2)(x – 3)(x – 6) = 0$$

$$(x – 2)(x^2 – 9x + 18) = 0$$

$$x^3 – 9x^2 + 18x – 2x^2 + 18x – 36 = 0$$

$$x^3 – 11x^2 + 36x – 36 = 0$$

---

VSAQ-11 : If 1,2,3 and 4 are the roots of x⁴+ax³+bx²+cx+d=0, then find the values of a,b,c and d

Given the roots are $$\alpha = 1$$ $$\beta = 2$$ $$\gamma = 3$$ $$\delta = 4$$

$$a = a_1 = -s_1 = -(\alpha + \beta + \gamma + \delta) = -(1 + 2 + 3 + 4) = -10$$

$$b = a_2 = s_2 = (\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta) = (1\cdot2 + 1\cdot3 + 1\cdot4 + 2\cdot3 + 2\cdot4 + 3\cdot4) = (2 + 3 + 4 + 6 + 8 + 12) = 35$$

$$c = a_3 = -s_3 = -(\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta) = -(1\cdot2\cdot3 + 1\cdot2\cdot4 + 1\cdot3\cdot4 + 2\cdot3\cdot4) = -(6 + 8 + 12 + 24) = -50$$

$$d = a_4 = s_4 = \alpha\beta\gamma\delta = (1)(2)(3)(4) = 24$$