## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1

Question 1.

If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.

Solution:

Let the side of cubic box be x.

They its volume is x³

Sides are by 1, 2, 3 units respectively.

Then the volume of the new box is (x + 1) (x + 2) (x + 3).

∴ Increase in volume is

(x + 1) (x + 2) (x + 3) – x³ = 52 (given)

x³ + 6x² + 11x + 6 – x³ = 52

6x² + 11x – 46 = 0

(6x + 23) (x – 2) = 0

x = – \(\frac{23}{6}\) or = 2

x = – \(\frac{23}{6}\) is not admissible ∴ x – 2

Hence volume of the old cubic box = 8 cubic units

That of new box = 60 cubic units.

Samacheer Kalvi 12th Maths Guide

Question 2.

Construct a cubic equation with roots

Solution:

(i) 1, 2, and 3

The roots are 1, 2 and 3

∴ Equation is

(x – 1)(x – 2)(x – 3) = 0

x³ – 6x² + 11x – 6 = 0

(ii) 1, 1, and -2

The roots are 1, 1 and -2

The Equation is

(x – 1)(x – 1)(x + 2) = 0

x³ – 3x + 2 = 0

(iii) 2, -2, and 4.

The roots are 2, -2 and 4

The Equation is

(x – 2)(x + 2)(x – 4) = 0

(x^{4} – 4)(x – 4) = 0

x³ – 4x² – 4x + 16 = 0

Question 3.

If α, β and γ are the roots of the cubic equation x³ + 2x² + 3x + 4 = 0, form a cubic equation whose roots are

(i) 2α, 2β, 2γ,

(ii) \(\frac{1}{α}\), \(\frac{1}{β}\), \(\frac{1}{γ}\)

(iii) – α, – β, – γ

Solution:

x³ + 2x² + 3x + 4 = 0

a, P, y are the roots of their equation

α + β + γ = -2

αβ + βγ + γα = 3

αβγ = -4

(i) The new roots are 2α, 2β, 2γ

(a) 2α + 2β + 2γ = 2(α + β + γ) = 2(-2) = 4

(b) (2α) (2β) + (2β) (2γ) + (2γ) (2α)

= 4[αβ + βγ + γα] = 4(3) = 12

(c) (2α) (2β) (2γ) – 8(αβγ) = 8(-4) = -32

∴ The required equation is

x³ + 4x² + 12x + 32 = 0

(ii) The new roots are \(\frac{1}{α}\), \(\frac{1}{β}\), \(\frac{1}{γ}\)

The required equation is

x³+ \(\frac{3}{4}\)x² + \(\frac{2}{4}\)x + \(\frac{1}{4}\) = 0

4x³ + 3x² + 2x + 1 = 0

(iii) The new roots are -α, -β, -γ

(a) -α – β – γ = -(-2) = 2

(b) (-α) (-β) + -(β) (-γ) + (-γ) (-α)

= αβ + βγ + αγ = 3

(c) (-α) (-β) (-γ) = -αβγ = -(-4) = 4

The required equation is

x³ – 2x³ + 3x – 4 = 0

Question 4.

Solve the equation 3x³ – 16x² + 23x – 6 = 0 if the product of two roots is 1.

Solution:

Given that the product of two roots is 1

Let us assume that the roots be α \(\frac{1}{α}\) and β

⇒ β = 2

Hence 2 is a root of the equation ⇒ (x – 2) is a functions.

x – 2) 3x³ – 16x² + 23x – 6 (3x² -10x + 3

3x² – 10x + 3 = 0

(3x – 1)(x – 3) = 0

⇒ x = \(\frac{1}{3}\), 3

Hence the roots are 3, \(\frac{1}{3}\) and 2.

Question 5.

Find the sum of squares of roots of the equation 2x^{4} – 8x³ + 6x² – 3 = 0.

Solution:

Let the roots be α, β, γ, δ

α + β + γ + δ = -(-\(\frac{8}{2}\)) = 4

αβ + βy + γδ + αγ + αδ + βδ = \(\frac{6}{2}\) = 3

αβγ + αβδ + βγδ + αγδ = \(\frac{0}{2}\) = 0

αβyδ = –\(\frac{3}{2}\)

To find α2+ β2 + γ2+ δ2

= (α + β + γ + δ)2 – 2(α + βγ+….)

= (4)² – 2(3) = 16 – 6 = 10

Question 6.

Solve the equation x³ – 9x² + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3 : 2.

Solution:

Given that two of its roots are in the ratio 3 : 2

Let us assume that the roots are 3α, 2α and β

3α + 2α+ β = -(-9) = 9

⇒ 5α + β = 9 ……… (1)

(3α) (2α) + (2α) (β) + (β) (3α) = 14

⇒ 6α² + 5αβ = 14 ……… (3)

From (3) ⇒

Substitute in (1)

5α – \(\frac{4}{α^2}\)

5α³ – 9α² – 4 = 0

(α – 2) is a factor.

5α²+ α + 2 = 0 gives imaginary roots

When α = 2, substitute in (1)

⇒ 5.(2) + β = 9

⇒ β = -1.

Hence the roots are 6, 4, -1

(∵ 3α, 2α, β) are the roots.

Question 7.

If α, β and γ are the roots of the polynomial equation ax³ + bx² + cx + d = 0, find the value of Σ\(\frac{α}{βγ}\) terms of the coefficients.

Solution:

Let α, β, γ are the roots.

∴ α + β + γ = –\(\frac{b}{a}\)

Question 8.

If α, β, γ and δ are the roots of the polynomial equation 2x^{4} + 5x³ – 7x² + 8 = 0, find a quadratic equation with integer coefficients whose roots are α + β + γ + δ and αβγδ.

Solution:

α, β, γ and δ be the roots then

α + β + γ + δ = –\(\frac{5}{2}\)

αβ + βγ + γδ + αδ + βδ + αγ = –\(\frac{7}{2}\)

αβγ + αγδ + αβδ + βγδ = 0

αβγδ = –\(\frac{8}{2}\) = 4

Now the new roots are (α + β + γ + δ) and (αβγδ) i.e., –\(\frac{5}{2}\) and 4

∴ Sum of the roots = –\(\frac{5}{2}\) + 4 = \(\frac{3}{2}\)

Product of the roots = (-\(\frac{5}{2}\))(4) = -10

Required equation is

x² – \(\frac{3x}{2}\) – 10 = 0

2x² – 3x – 20 = 0

Question 9.

If p and q are the roots of the equationlx² + nx + n = 0,

show that,

Solution:

lx² + nx + n = 0 ⇒ p and q are the roots.

Question 10.

If the equations x² + px + q = 0 and x² + p’x + q’ = 0 have a common root, show

that it must be equal to \(\frac{pq’-p’q}{q-q’}\) and \(\frac{q-q’}{p’-p}\)

Solution:

Let it be α then

α² + pα + q = 0

α² + p’α + q’ = 0

Solving by cross multiplication method, we have

Question 11.

Formulate into a mathematical problem to find a number such that when its cube root is added to it, the result is 6.

Solution:

Let the number be x.

\(\sqrt[3]{x}\) +x = 6

x^{\(\frac[1]{3}\)} = 6 – x

along x = (6 – x)³

= 216 – 108x + 18x² – x³

⇒ x³ – 18x² + 109x – 216 = 0

Question 12.

A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.

Solution:

Let the length of the tree cut away = x

the length of the free left = y

Given x + y = 12 …….. (1)

\(\sqrt[3]{x}\) = y …….. (2)

From (2) y³ = x

From (1 x = 12 – y

we have y³ = 12 – y

y³ + y + -12 = 0

change y to x ⇒ x³ + x – 12 = 0