Unlock the secrets of polynomial equations with these magical solutions!
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
1. Solve the cubic equation: 2x³ - x² - 18x + 9 = 0 if sum of two of its roots vanishes.
Step 1: Let the roots be α, β, γ with the condition that α + β = 0.
Step 2: From Vieta's formulas for the cubic equation ax³ + bx² + cx + d = 0:
• Sum of roots: α + β + γ = -b/a = 1/2
• Sum of product of roots two at a time: αβ + αγ + βγ = c/a = -18/2 = -9
• Product of roots: αβγ = -d/a = -9/2
Step 3: Since α + β = 0, from sum of roots:
0 + γ = 1/2 ⇒ γ = 1/2
Step 4: From sum of product of roots:
αβ + α(1/2) + β(1/2) = -9
But α + β = 0, so:
αβ + (1/2)(α + β) = αβ = -9
Step 5: From product of roots:
αβγ = (-9)(1/2) = -9/2 which matches -d/a
Step 6: Now we know two numbers (α and β) that add to 0 and multiply to -9.
They satisfy x² - (α+β)x + αβ = x² + 0x - 9 = x² - 9 = 0
So α = 3, β = -3
The roots are:
3
-3
½
2. Solve the equation 9x³ - 36x² + 44x - 16 = 0 if the roots form an arithmetic progression.
Step 1: Let the roots be in arithmetic progression: a - d, a, a + d
Step 2: From Vieta's formula for sum of roots:
(a - d) + a + (a + d) = 3a = 36/9 = 4 ⇒ a = 4/3
Step 3: From sum of product of roots two at a time:
(a-d)a + a(a+d) + (a-d)(a+d) = 3a² - d² = 44/9
Substitute a = 4/3:
3(16/9) - d² = 44/9 ⇒ 48/9 - d² = 44/9 ⇒ d² = 4/9 ⇒ d = ±2/3
Step 4: The roots are:
When d = 2/3: (4/3 - 2/3, 4/3, 4/3 + 2/3) = (2/3, 4/3, 2)
When d = -2/3: (4/3 + 2/3, 4/3, 4/3 - 2/3) = (2, 4/3, 2/3)
The roots are:
⅔
⁴⁄₃
2
3. Solve the equation 3x³ - 26x² + 52x - 24 = 0 if its roots form a geometric progression.
Step 1: Let the roots be in geometric progression: a/r, a, ar
Step 2: From Vieta's formula for product of roots:
(a/r) × a × ar = a³ = 24/3 = 8 ⇒ a = 2
Step 3: From sum of roots:
2/r + 2 + 2r = 26/3
Multiply by 3r:
6 + 6r + 6r² = 26r ⇒ 6r² - 20r + 6 = 0 ⇒ 3r² - 10r + 3 = 0
Step 4: Solve the quadratic:
r = [10 ± √(100 - 36)]/6 = [10 ± 8]/6
So r = 3 or r = 1/3
Step 5: The roots are:
For r = 3: (2/3, 2, 6)
For r = 1/3: (6, 2, 2/3)
The roots are:
⅔
2
6
4. Determine k and solve the equation 2x³ - 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
Step 1: Let the roots be α, β, γ with γ = 2(α + β)
Step 2: From Vieta's formula for sum of roots:
α + β + γ = α + β + 2(α + β) = 3(α + β) = 6/2 = 3 ⇒ α + β = 1 ⇒ γ = 2
Step 3: Since γ = 2 is a root, substitute x = 2 into the equation:
2(8) - 6(4) + 3(2) + k = 0 ⇒ 16 - 24 + 6 + k = 0 ⇒ k = 2
Step 4: Now we can factor out (x - 2):
2x³ - 6x² + 3x + 2 = (x - 2)(2x² - 2x - 1)
Step 5: Solve the quadratic:
x = [2 ± √(4 + 8)]/4 = [2 ± √12]/4 = [1 ± √3]/2
The roots are:
(1 - √3)/2 ≈ -0.366
(1 + √3)/2 ≈ 1.366
2
Value of k:2
5. Find all zeros of the polynomial x⁶ - 3x⁵ - 5x⁴ + 22x³ - 39x² - 39x + 135, if it is known that
1 + 2i and √3 are two of its zeros.
Step 1: Since coefficients are real, complex roots come in conjugate pairs.
If 1 + 2i is a root, then 1 - 2i is also a root.
Step 3: For √3 to be a root, and since coefficients are rational, -√3 must also be a root.
Find quadratic factor for irrational roots:
(x - √3)(x + √3) = x² - 3