Theory Of Equations (VSAQs)

Maths-2A | 4. Theory Of Equations – VSAQs:
Welcome to VSAQs in Chapter 4: Theory Of Equations. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

---

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 $$\alpha = 1$$ $$\beta = 2$$ $$\gamma = 3$$ $$\delta = 4$$

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

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

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

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