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

**Exercise 3.1 Class 12 Maths State Board 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 the cube be ‘x’

Sides of cuboid are (x + 1) (x + 2) (x + 3)

∴ Volume of cuboid = x^{3} + 52

⇒ (x + 1) (x + 2) (x + 3) = x^{3} + 52

⇒ (x^{2} + 3x + 2)(x + 3) = x^{3} + 52

⇒ x^{3} + 3x^{2} + 3x^{2} + 9x + 2x + 6 – x^{3} – 52 = 0

⇒ 6x^{2} + 11x – 46 = 0 (÷2)

⇒ (x – 2) (6x + 23) = 0

⇒ x – 2 = 0 or 6x + 23 = 0

⇒ x = 2 or x = \(-\frac{23}{6}\) (not possible)

∴ x = 2

Volume of cube = 2^{3} = 8

Volume of cuboid = 52 + 8 = 60 cubic units

**12th Maths Exercise 3.1 Question 2.**

Construct a cubic equation with roots

(i) 1, 2 and 3

(ii) 1, 1 and -2

(iii) 2, \(\frac { 1 }{ 2 }\) and 1

Solution:

(i) Given roots are α = 1, β = 2, γ = 3

The cubic equation is

x^{3} – x^{2} (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0

⇒ x^{3} – x^{2} (1 + 2 + 3) + x (2 + 6 + 3) – (1) (2) (3) = 0

⇒ x^{3} – 6x^{2} + 11x – 6 = 0

(ii) α = 1, β = 1, γ = -2

The cubic equation is

x^{3} – x^{2} (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0

⇒ x^{3} – x^{2} (1 + 1 – 2) + x (1 – 2 – 2) – (1) (1) (-2) = 0

⇒ x^{3} – 0x^{2} – 3x + 2 = 0

⇒ x^{3} – 3x + 2 = 0

(iii) α = 2, β = \(\frac { 1 }{ 2 }\), γ = 1

The cubic equation is

x^{3} – x^{2} (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0

2x^{3} – 7x^{2} + 7x – 2 = 0

**12th Maths Theory Of Equations Question 3.**

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

(i) 2α, 2β, 2γ

(ii) \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)

(iii) -α, -β, -γ

Solution:

(i) Given that α, β, γ are the roots of x^{3} + 2x^{2} + 3x + 4 = 0

Compare with x^{3} + bx^{2} + cx + d = 0

b = 2, c = 3, d = 4

α + β + γ = -6 = -2

αβ + βγ + γα = c = 3

αβγ = -d = -4

Given roots are 2α, 2β, 2γ

2α + 2β + 2γ = 2 (α + β + γ)

= 2 (-2)

= -4

(2α) (2β) + (2β) (2γ) + (2γ) (2α) = (4αβ + 4βγ + 4γα)

= 4(αβ + βγ + γα)

= 4(3)

= 12

(2α) (2β) (2γ) = 8(αβγ)

= 8(-4)

= -32

The equation is

x^{3} – x^{2} (2α + 2β + 2γ) + x (4αβ + 4βγ + 4γα) – 8 (αβγ) = 0

⇒ x^{3} – x^{2} (-4) + x (12) – (-32) = 0

⇒ x^{3} + 4x^{2} + 12x + 32 = 0

(ii) The given roots are \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)

The cubic equation is

4x^{3} + 3x^{2} + 2x + 1 = 0 (Multiply by 4)

(iii) The given roots are -α, -β, -γ

The cubic equation is

x^{3} – x^{2} (-α – β – γ) + x (αβ + βγ + γα) + (αβγ) = 0

⇒ x^{3} + x^{2} (α + β + γ) + x (αβ + βγ + γα) + (αβγ) = 0

⇒ x^{3} + x^{2} (-2) + x (3) – 4 = 0

⇒ x^{3} – 2x^{2} + 3x – 4 = 0

**12th Maths Chapter 3 Exercise 3.1 Question 4.**

Solve the equation 3x^{3} – 16x^{2} + 23x – 6 = 0 if the product of two roots is 1.

Solution:

The given equation is 3x^{3} – 16x^{2} + 23x – 6 = 0

⇒ \(x^{3}-\frac{16}{3} x^{2}+\frac{23}{3} x-2=0\) (÷3)

Let the roots be α, β, γ

α + β + γ = -b = \(\frac{16}{3}\) …….. (1)

αβ + βγ + γα = c = \(\frac{23}{3}\) …….. (2)

αβγ = -d = 2 ……. (3)

Given that αβ = 1

from (3), γ = 2

Substitute \(\beta=\frac{1}{\alpha}\), γ = 2 in (1)

⇒ \(\alpha+\frac{1}{\alpha}+2=\frac{16}{3}\)

\(\Rightarrow \frac{\alpha^{2}+1}{\alpha}=\frac{16}{3}-2\)

⇒ \(\frac{\alpha^{2}+1}{\alpha}=\frac{10}{3}\)

⇒ 3α^{2} + 3 = 10α

⇒ 3α^{2} – 10α + 3 = 0

⇒ (3α – 1) (α – 3) = 0

⇒ α = \(\frac{1}{3}\), 3

α = \(\frac{1}{3}\), β = 3 (or) when a = 3, β = \(\frac{1}{3}\)

∴ The roots are 3, \(\frac{1}{3}\), 2

(or) when γ = 2, by synthetic division method.

The factors are (x – 2) (x – 3) (3x – 1)

∴ The roots are 2, 3, \(\frac{1}{3}\)

**12th Maths 3.1 Question 5.**

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

Solution:

The given equation is 2x^{4} – 8x^{3} + 6x^{2} – 3 = 0.

(÷ 2) ⇒ x^{4} – 4x^{3} + 3x^{2} – \(\frac{3}{2}\) = 0

Let the roots be α, β, γ, δ

α + β + γ + δ = -b = 4

(αβ + βγ + γδ + αδ + αγ + βδ) = c = 3

αβγ + βγδ + γδα = -d = 0

αβγδ = \(\frac{-3}{2}\)

To Find α^{2} + β^{2} + γ^{2} + δ^{2} = (α + β + γ + δ)^{2} – 2 (αβ + βγ + γδ + αδ + αγ + βδ)

= (4)^{2} – 2(3)

= 16 – 6

= 10

**12 Maths Exercise 3.1 Question 6.**

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

Solution:

The given equation is x^{3} – 9x^{2} +14x + 24 = 0.

Since the two roots are in the ratio 3 : 2.

The roots are α, 3λ, 2λ

α + 3λ + 2λ = -b = 9

⇒ α + 5λ = 9 …… (1)

(α) (3λ) (2λ) = -24

6λ^{2}α = -24

⇒ λ^{2}α = -4 …… (2)

(1) ⇒ α = 9 – 5λ

(2) ⇒ λ^{2} (9 – 5λ) = -4

9λ^{2} – 5λ^{3} + 4 = 0

5λ^{3} – 9λ^{2} – 4 = 0

(λ – 2) (5λ^{2} + λ + 2) = 0

λ = 2, 5λ^{2} + λ + 2 = 0 has only Imaginary roots Δ < 0

when λ = 2, α = 9 – 5 (2) = 9 – 10 = -1

The roots are α, 3λ, 2λ i.e., -1, 6, 4

**Samacheer Kalvi Guru 12th Maths Question 7.**

If α, β and γ are the roots of the polynomial equation ax^{3} + bx^{2} + cx + d= 0, find the value of \(\Sigma \frac{\alpha}{\beta \gamma}\) in terms of the coefficients.

Solution:

The given equation is ax^{3} + bx^{2} + cx + d = 0.

÷a ⇒ \(x^{3}+\frac{b}{a} x^{2}+\frac{c}{a} x+\frac{d}{a}=0\)

Let the roots be α, β, γ

α + β + γ = \(-\frac{b}{a}\)

αβ + βγ + γα = \(\frac{c}{a}\)

αβγ = \(-\frac{d}{a}\)

To find:

**Samacheer Kalvi 12 Maths Solutions Question 8.**

If α, β, γ and δ are the roots of the polynomial equation 2x^{4} + 5x^{3} – 7x^{2} + 8 = 0, find a quadratic equation with integer coefficients whose roots are α + β + γ + δ and αβγδ.

Solution:

The given equation is 2x^{4} + 5x^{3} – 7x^{2} + 8 = 0.

÷ 2 ⇒ \(x^{4}+\frac{5}{2} x^{3}-\frac{7}{2} x^{2}+4=0\)

Let the roots be α, β, γ, δ

α + β + γ + δ = \(-\frac{5}{2}\)

αβγδ = -4

To form the quadratic equation with the given roots α + β + γ + δ, αβγδ.

x^{2} – x(S.O.R) + P.O.R = 0

\(x^{2}-x\left(\frac{-5}{2}-4\right)+\left(\frac{-5}{2}\right)(-4)=0\)

\(\Rightarrow x^{2}-x\left(\frac{-13}{2}\right)+10=0\)

2x^{2} + 13x + 20 = 0

**Samacheer Kalvi 12th Maths Solutions Question 9.**

If p and q are the roots of the equation lx^{2} + nx + n = 0, show that \(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)

Solution:

The given equation is lx^{2} + nx + n = 0.

p + q = \(-\frac{n}{l}\), pq = \(\frac{n}{l}\)

**Samacheer Kalvi.Guru 12th Maths Question 10.**

If the equations x^{2} + px + q = 0 and x^{2} + p’x + q’ = 0 have a common root, show that it must be equal to \(\frac{p q^{\prime}-p^{\prime} q}{q-q^{\prime}}\) or \(\frac{q-q^{\prime}}{p^{\prime}-p}\)

Solution:

If α is the common root, then.

α^{2} + pα + q = 0 ……. (1)

α^{2} + p’α + q’ = 0 ……… (2)

Subtracting α (p – p’) = q’ – q

\(\alpha=\frac{q^{\prime}-q}{p-p^{\prime}}=\frac{q-q^{\prime}}{p^{\prime}-p}\) …….. (3)

Eliminating α from (1) & (2)

**Samacheer Kalvi Class 12 Maths Solutions 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.

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

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

Cubing on both sides

x = (6 – x)^{3}

⇒ x = 216 – 3 (6)^{2} (x) + 3(6) (x)^{2} – x^{3}

⇒ x = 216 – 108x + 18x^{2} – x^{3}

⇒ x^{3} – 18x^{2} + 109x – 216 = 0

**Samacheerkalvi.Guru 12th Maths 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 two parts be x and (12 – x)

Given that \(x=\sqrt[3]{12-x}\)

Cubing on both side,

x^{3} = 12 – x

⇒ x^{3} + x – 12 = 0

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

**Samacheer Kalvi 12th Maths Example Sums Question 1.**

Construct a cubic equation with roots 2, 3, 4.

Solution:

Given roots are 2, 3, 4

Take α = 1; β = 3; γ = 4

The required cubic polynomial is

x^{3} – (α + β + γ) x^{2} + (αβ + βγ + γα) x – αβγ = 0

x^{3} – (1 + 3 + 4)x^{2} + (3 + 12 + 4) x – 12 = 0

x^{3} – 8x^{2} + 19 x – 12 = 0

**Samacheer Kalvi 12th Maths Question 2.**

If α, β, γ are the roots of the cubic equation x^{3} – 6x^{2} + 11x – 6 = 0. From a cubic equation whose roots are 2α, 2β, 2γ.

Solution:

Given that α, β, γ are the roots of x^{3} – 6x^{2} + 11x – 6 = 0 … α, β, γ

α + β + γ = 6 …(1)

αβ + βγ + γα = 11 …(2)

αβγ = 6 …(3)

Form a cubic equation whose roots are 2α, 2β, 2γ.

∴ 2α + 2β + 2γ = 2(α + β + γ) = 2(6) = 12

The required cubic equation is

x^{3} – (2α + 2β + 2γ)x^{2} + (4αβ + 4βγ + 4γα) x – (2α) (2β) (2γ) = 0

x^{3} – 12x^{2} + 44x – 48 = 0

**Samacheer Kalvi Maths 12th Question 3.**

If the roots of x^{4} + 5x^{3} – 30x^{2} – 40x + 64 = 0 are in G.P; then find the roots.

Solution:

**Samacheer Kalvi 12th Maths Guide Question 4.**

Determine the value of k such that the equation (2k – 5)x^{2} – 4x – 15 = 0 and (3k – 8)x^{2} – 5x – 21 = 0 may have a common root.

Solution:

If α be the common root, the two equations.

(2k – 5) α^{2} – 4α – 15 = 0

(3k – 8) α^{2} – 5α – 21 = 0 These are the linear equation is α^{2} and α.

By cross multiplication rule

**12th Maths Samacheer Kalvi Question 5.**

If α, β, γ are the roots of the equation x^{3} + px^{2} + qx +1 = 0. Find the value of the following in terms of coefficients.

Solution:

**Samacheer Kalvi Guru 12 Maths Question 6.**

Solution: