✨ Complete Math Magic ✨

All Cubic Equations & Polynomial Problems Solved

Exercise 3.1

Problem 1: The Expanding Box

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.

Let the original side of the cube be x units.

Original volume = x³ cubic units.

New dimensions: (x+1), (x+2), (x+3).

New volume = (x+1)(x+2)(x+3).

Volume increase = New volume - Original volume = 52.

So: (x+1)(x+2)(x+3) - x³ = 52.

Let's expand (x+1)(x+2)(x+3):

= (x² + 3x + 2)(x + 3)

= x³ + 3x² + 2x + 3x² + 9x + 6

= x³ + 6x² + 11x + 6

Now subtract x³: 6x² + 11x + 6 = 52

Simplify: 6x² + 11x - 46 = 0

Solve the quadratic equation:

x = [-11 ± √(121 + 1104)] / 12

x = [-11 ± √1225] / 12

x = [-11 ± 35] / 12

Positive solution: x = (24)/12 = 2

New volume = (2+1)(2+2)(2+3) = 3×4×5 = 60 cubic units.

🎉 Final Answer: 60 cubic units

Problem 2: Construct Cubic Equations

Construct a cubic equation with roots:
(i) 1, 2, and 3
(ii) 1, 1, and -2
(iii) 2, ½, and 1

Part (i): Roots 1, 2, 3

General form: (x - α)(x - β)(x - γ) = 0

Substitute roots: (x - 1)(x - 2)(x - 3) = 0

Multiply first two terms: (x² - 3x + 2)(x - 3) = 0

Final expansion: x³ - 6x² + 11x - 6 = 0

Part (ii): Roots 1, 1, -2

Equation: (x - 1)²(x + 2) = 0

Expand (x - 1)²: x² - 2x + 1

Multiply: (x² - 2x + 1)(x + 2) = 0

Final expansion: x³ + 0x² - 3x + 2 = 0 or x³ - 3x + 2 = 0

Part (iii): Roots 2, ½, 1

Equation: (x - 2)(x - ½)(x - 1) = 0

Multiply by 2 to eliminate fraction: (x - 2)(2x - 1)(x - 1) = 0

Multiply first two terms: (2x² - 5x + 2)(x - 1) = 0

Final expansion: 2x³ - 7x² + 7x - 2 = 0

🎉 Final Answers:
(i) x³ - 6x² + 11x - 6 = 0
(ii) x³ - 3x + 2 = 0
(iii) 2x³ - 7x² + 7x - 2 = 0

Problem 3: Transforming Roots

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) 1/α, 1/β, 1/γ
(iii) -α, -β, -γ

First, recall Vieta's formulas for original equation:

α + β + γ = -2 (sum of roots)

αβ + αγ + βγ = 3 (sum of products)

αβγ = -4 (product of roots)

Part (i): Roots 2α, 2β, 2γ

New sum: 2α + 2β + 2γ = 2(α+β+γ) = 2(-2) = -4

New sum of products: 4(αβ + αγ + βγ) = 4×3 = 12

New product: 8αβγ = 8×(-4) = -32

Thus equation: x³ + 4x² + 12x + 32 = 0

Part (ii): Roots 1/α, 1/β, 1/γ

Let y = 1/x, so x = 1/y

Substitute into original equation: (1/y)³ + 2(1/y)² + 3(1/y) + 4 = 0

Multiply by y³: 4y³ + 3y² + 2y + 1 = 0

Part (iii): Roots -α, -β, -γ

New sum: -α - β - γ = -(α+β+γ) = 2

New sum of products: (-α)(-β) + (-α)(-γ) + (-β)(-γ) = αβ + αγ + βγ = 3

New product: (-α)(-β)(-γ) = -αβγ = 4

Thus equation: x³ - 2x² + 3x - 4 = 0

🎉 Final Answers:
(i) x³ + 4x² + 12x + 32 = 0
(ii) 4x³ + 3x² + 2x + 1 = 0
(iii) x³ - 2x² + 3x - 4 = 0

Problem 4: Solve Cubic with Given Condition

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

Let the roots be α, β, γ with αβ = 1.

From Vieta's formulas for ax³ + bx² + cx + d = 0:

1. α + β + γ = -b/a = 16/3

2. αβ + αγ + βγ = c/a = 23/3

3. αβγ = -d/a = 2

Given αβ = 1, from 3rd formula: γ = 2.

Now from 1st formula: α + β = 16/3 - 2 = 10/3

We know αβ = 1 and α + β = 10/3.

These are roots of: x² - (10/3)x + 1 = 0

Multiply by 3: 3x² - 10x + 3 = 0

Solutions: x = [10 ± √(100 - 36)] / 6 = [10 ± 8]/6

So x = 3 or x = 1/3

Thus the roots are: 3, \frac{1}{3}, 2

🎉 Final Answer: Roots are 2, 3, and 1/3

Problem 5: Sum of Squares of Roots

Find the sum of squares of roots of the equation 2x⁴ - 8x³ + 6x² - 3 = 0.

Let the roots be α, β, γ, δ.

From Vieta's formulas for quartic ax⁴ + bx³ + cx² + dx + e = 0:

1. α + β + γ + δ = -b/a = 4

2. αβ + αγ + αδ + βγ + βδ + γδ = c/a = 3

We need to find α² + β² + γ² + δ².

Recall: (α+β+γ+δ)² = α²+β²+γ²+δ² + 2(αβ+αγ+αδ+βγ+βδ+γδ)

Thus: α²+β²+γ²+δ² = (α+β+γ+δ)² - 2(αβ+αγ+αδ+βγ+βδ+γδ)

Substitute values: = 4² - 2×3 = 16 - 6 = 10

🎉 Final Answer: Sum of squares of roots is 10

Additional Problems

Problem 6: Roots in Ratio

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

Let the roots be 3k, 2k, r (since two roots are in ratio 3:2).

From Vieta's formulas:

1. Sum: 3k + 2k + r = 9 ⇒ 5k + r = 9

2. Sum of products: 3k·2k + 3k·r + 2k·r = 14 ⇒ 6k² + 5kr = 14

3. Product: 3k·2k·r = -24 ⇒ 6k²r = -24 ⇒ k²r = -4

From 1st equation: r = 9 - 5k

Substitute into 3rd equation: k²(9 - 5k) = -4 ⇒ 9k² - 5k³ = -4

Rearrange: 5k³ - 9k² - 4 = 0

Try k=1: 5 - 9 - 4 = -8 ≠ 0

Try k=2: 40 - 36 - 4 = 0 ✓

Thus k=2 is a root.

Now find r: r = 9 - 5×2 = -1

So roots are: 3×2=6, 2×2=4, -1

🎉 Final Answer: Roots are -1, 4, and 6

Problem 7: Sum of Fractional Roots

If α, β, and γ are the roots of the polynomial equation ax³ + bx² + cx + d = 0, find the value of ∑(α/βγ) in terms of the coefficients.

We need to find α/βγ + β/αγ + γ/αβ.

Combine the fractions: = (α² + β² + γ²)/αβγ

From Vieta's formulas:

α + β + γ = -b/a

αβ + αγ + βγ = c/a

αβγ = -d/a

We know (α+β+γ)² = α² + β² + γ² + 2(αβ + αγ + βγ)

Thus α² + β² + γ² = (b²/a²) - 2(c/a)

Now our expression becomes:

[(b²/a²) - (2c/a)] / (-d/a) = [b² - 2ac] / (-a²d/a) = [b² - 2ac] / (-ad)

Simplify: = (2ac - b²)/ad

🎉 Final Answer: (2ac - b²)/ad

Problem 8: Quadratic from Quartic Roots

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.

From Vieta's formulas for quartic equation:

1. α + β + γ + δ = -b/a = -5/2

2. αβγδ = e/a = 8/2 = 4 (when equation is ax⁴ + bx³ + cx² + dx + e = 0)

We need quadratic with roots -5/2 and 4.

Sum of roots: -5/2 + 4 = 3/2

Product of roots: (-5/2)×4 = -10

Thus equation: x² - (3/2)x - 10 = 0

Multiply by 2 for integer coefficients: 2x² - 3x - 20 = 0

🎉 Final Answer: 2x² - 3x - 20 = 0

Problem 9: Roots Relationship

If p and q are the roots of the equation lx² + nx + n = 0, show that √(p/q) + √(q/p) + √(n/l) = 0.

From Vieta's formulas:

1. p + q = -n/l

2. pq = n/l

Consider the expression √(p/q) + √(q/p):

Combine terms: = (p + q)/√(pq)

Substitute from Vieta: = (-n/l)/√(n/l) = -√(n/l)

Now add the third term: -√(n/l) + √(n/l) = 0

🎉 Final Answer: Proved the given relationship

Problem 10: Common Root

If the equations x² + px + q = 0 and x² + p'x + q' = 0 have a common root, show that it must be equal to (pq' - p'q)/(q - q') or (q - q')/(p' - p).

Let α be the common root.

Then: α² + pα + q = 0 and α² + p'α + q' = 0

Subtract the equations: (p - p')α + (q - q') = 0

Solve for α: α = (q' - q)/(p - p') = (q - q')/(p' - p)

Now from first equation: α² = -pα - q

Multiply second equation by q and first by q':

qα² + qp'α + qq' = 0

q'α² + q'pα + qq' = 0

Subtract: (q - q')α² + (qp' - q'p)α = 0

Factor out α: α[(q - q')α + (qp' - q'p)] = 0

For non-zero α: α = (q'p - qp')/(q - q') = (pq' - p'q)/(q - q')

🎉 Final Answer: Proved both forms of the common root

Problem 11: The Broken Tree

A 12 meter 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. Find the height of the part which was left standing.

Tree Height = 12m

Let x = height of standing part (in meters).

Then fallen part length = 12 - x.

According to problem: x = ∛(12 - x).

Cube both sides: x³ = 12 - x.

Rearrange: x³ + x - 12 = 0.

Try x=2: 8 + 2 - 12 = -2 ≠ 0.

Try x=2.5: 15.625 + 2.5 - 12 = 6.125 ≠ 0.

Try x=2.2: 10.648 + 2.2 - 12 = 0.848 ≈ 0 (close).

For exact solution, we can use numerical methods.

The exact solution is approximately x ≈ 2.1544 meters.

🎉 Final Answer: Height of standing part ≈ 2.15 meters