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

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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 the cubic box are x, x, x.
Then its volume is x³
Sides are by 1, 2, 3 units respectively.
Cuboid sides be (x + 1) (x + 2) (x + 3).
volume of cuboid is
⇒ (x + 1) (x + 2) (x + 3) = x³ + 52
⇒ (x² + 3x + 2) (x + 3) = x³ + 52
⇒ x³ + 3x² + 2x + 3x² + 9x + 6 = x³ + 52
⇒ 6x² + 11x + 6 – 52 = 0
⇒ 6x² + 11x – 46 = 0
⇒ 6x² – 12x + 23x – 46 = 0
⇒ 6x (x – 2) + 23(x – 2) = 0
⇒ x – 2 = 0, 6x + 23 = 0
⇒ x = 2, 6x = -23
∴ x = 2, x = \(\frac { -23 }{ 6 }\) (is not possible)
volume of cube = x³ = 2³ = 8
Volume of cuboid = 52 + 8 = 60

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 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
x3 – x2 (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
⇒ x3 – x2 (1 + 2 + 3) + x (2 + 6 + 3) – (1) (2) (3) = 0
⇒ x3 – 6x2 + 11x – 6 = 0
(ii) α = 1, β = 1, γ = -2
The cubic equation is
x3 – x2 (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
⇒ x3 – x2 (1 + 1 – 2) + x (1 – 2 – 2) – (1) (1) (-2) = 0
⇒ x3 – 0x2 – 3x + 2 = 0
⇒ x3 – 3x + 2 = 0
(iii) α = 2, β = \(\frac { 1 }{ 2 }\), γ = 1
The cubic equation is
x3 – x2 (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q2
2x3 – 7x2 + 7x – 2 = 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) \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
(iii) -α, -β, -γ
Solution:
(i) 2α, 2β, 2γ
x³ + 2x² + 3x + 4 = 0
∴ a = 1, b = 2, c = 3, d = 4
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-2}{1}\) = -2
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{3}{1}\) = 3
αβγ = \(\frac{-d}{a}\) = -4
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
x³ -(-4)x² + 12x – (-32) = 0
x³ + 4x² + 12x + 32 = 0
(ii) The given roots are \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
The cubic equation is
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q3
4x3 + 3x2 + 2x + 1 = 0 (Multiply by 4)
(iii) The new roots are -α, -β, -γ
\(\sum_{1}\) = -(α + β + γ) = -(-2) = 2
\(\sum_{2}\) = αβ + βγ + γα = 3
\(\sum_{3}\) = -(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:
The given equation is 3x3 – 16x2 + 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)
⇒ \(\alpha+\frac{1}{\alpha}+2=\frac{16}{3}\)
\(\Rightarrow \frac{\alpha^{2}+1}{\alpha}=\frac{16}{3}-2\)
⇒ \(\frac{\alpha^{2}+1}{\alpha}=\frac{10}{3}\)
⇒ 3α2 + 3 = 10α
⇒ 3α2 – 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.
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q4
The factors are (x – 2) (x – 3) (3x – 1)
∴ The roots are 2, 3, \(\frac{1}{3}\)

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

Question 5.
Find the sum of squares of roots of the equation 2x4 – 8x3 + 6x2 – 3 = 0.
Solution:
Let the roots be α, β, γ, δ
α + β + γ + δ = \(\frac{-b}{a}\) = \(\frac{8}{2}\) = 4
αβ + βy + γα + δα + βδ + γδ = \(\frac{c}{a}\) = \(\frac{6}{2}\) = 3
(a + b + c + d)² = a² + b² + c² + d² +2(ab + ac + ad + bc + bd + cd)
To find α²+ β² + γ²+ δ² = (α + β + γ + δ)² – 2(αβ + βy + γα + δα + βδ + γδ)
= (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:
The given equation is x3 – 9x2 +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
2α = -24
⇒ λ2α = -4 …… (2)
(1) ⇒ α = 9 – 5λ
(2) ⇒ λ2 (9 – 5λ) = -4
2 – 5λ3 + 4 = 0
3 – 9λ2 – 4 = 0
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q6
(λ – 2) (5λ2 + λ + 2) = 0
λ = 2, 5λ2 + 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

Question 7.
If α, β and γ are the roots of the polynomial equation ax3 + bx2 + cx + d= 0, find the value of \(\Sigma \frac{\alpha}{\beta \gamma}\) in terms of the coefficients.
Solution:
The given equation is ax3 + bx2 + 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 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q7

Question 8.
If α, β, γ and δ are the roots of the polynomial equation 2x4 + 5x3 – 7x2 + 8 = 0, find a quadratic equation with integer coefficients whose roots are α + β + γ + δ and αβγδ.
Solution:
The given equation is 2x4 + 5x3 – 7x2 + 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.
x2 – x(S.O.R) + P.O.R = 0
\(x^{2}-x\left(\frac{-5}{2}-4\right)+\left(\frac{-5}{2}\right)(-4)=0\)
\(\Rightarrow x^{2}-x\left(\frac{-13}{2}\right)+10=0\)
2x2 + 13x + 20 = 0

Question 9.
If p and q are the roots of the equation lx2 + nx + n = 0, show that \(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)
Solution:
The given equation is lx2 + nx + n = 0.
p + q = \(-\frac{n}{l}\), pq = \(\frac{n}{l}\)
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q9

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 \(\frac{p q^{\prime}-p^{\prime} q}{q-q^{\prime}}\) or \(\frac{q-q^{\prime}}{p^{\prime}-p}\)
Solution:
If α is the common root, then.
α2 + pα + q = 0 ……. (1)
α2 + 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 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q10
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Q10.1

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

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 – x3
⇒ x = 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 two parts be x and (12 – x)
Given that \(x=\sqrt[3]{12-x}\)
Cubing on both side,
x3 = 12 – x
⇒ x3 + x – 12 = 0

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

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
x3 – (α + β + γ) x2 + (αβ + βγ + γα) x – αβγ = 0
x3 – (1 + 3 + 4)x2 + (3 + 12 + 4) x – 12 = 0
x3 – 8x2 + 19 x – 12 = 0

Question 2.
If α, β, γ are the roots of the cubic equation x3 – 6x2 + 11x – 6 = 0. From a cubic equation whose roots are 2α, 2β, 2γ.
Solution:
Given that α, β, γ are the roots of x3 – 6x2 + 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
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 4
The required cubic equation is
x3 – (2α + 2β + 2γ)x2 + (4αβ + 4βγ + 4γα) x – (2α) (2β) (2γ) = 0
x3 – 12x2 + 44x – 48 = 0

Question 3.
If the roots of x4 + 5x3 – 30x2 – 40x + 64 = 0 are in G.P; then find the roots.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 5

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

Question 4.
Determine the value of k such that the equation (2k – 5)x2 – 4x – 15 = 0 and (3k – 8)x2 – 5x – 21 = 0 may have a common root.
Solution:
If α be the common root, the two equations.
(2k – 5) α2 – 4α – 15 = 0
(3k – 8) α2 – 5α – 21 = 0 These are the linear equation is α2 and α.
By cross multiplication rule
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 6

Question 5.
If α, β, γ are the roots of the equation x3 + px2 + qx +1 = 0. Find the value of the following in terms of coefficients.
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 7
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 8
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 9

Question 6.
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 10
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 11

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