Solve the equation 6x⁴ - 29x³ + 40x² - 7x - 12 = 0 given that product of two the roots is 2.
Answers
Answer:
Roots are 1 ± √2, 4/3 and 3/2
Step-by-step explanation:
Given Equation
6x⁴ - 29x³ + 40x² - 7x - 12 = 0
Comparing the equation with ax⁴ + bx³ + cx² + dx + e = 0
- a = 6
- b = - 29
- c = 40
- d = - 7
- e = - 12
Let the roots of the equation be α, β, γ, δ
Product of roots = e/a
⇒ αβγδ = - 12/6 = - 2
Given
Product of two roots = 2
⇒ αβ = 2
αβγδ = - 2
⇒ 2γδ = - 2
⇒ γδ = - 1
Let sum of α, β be p
Let's form the equation for α, β
⇒ x² - px + 2 = 0
Let sum of γ, δ be q
Again form the equationfor δ, γ
⇒ x² + qx - 1 = 0
For our convenience let's write the equation
⇒ rx² + qx - r = 0
[ since product = - r/r = - 1 ]
Multiplying the formed equations
⇒ ( x² + px + 2 )( rx² + qx - r ) = 0
⇒rx⁴ + qx³ - rx² + prx³ + pqx² - prx + 2rx² + 2qx - 2r = 0
⇒rx⁴ + ( q + pr )x³ + ( - r + pq + 2r )x² + ( - pr + 2q )x - 2r = 0
⇒ rx⁴ + ( q + pr )x³ + ( r + pq )x² + ( - pr + 2q )x - 2r = 0
Comparing the above equation with given Equation
- r = 6
- q + pr = - 29
- r + pq = 40
- 2q - pr = - 7
- - 2r = - 12
q + pr = - 29
⇒ q + 6p = - 29 -- eq(1)
pr - 2q = - 7
⇒ 2q - 6q = - 7 ---eq(2)
Adding EQ(1) and EQ( 2)
⇒ 3q = - 36
⇒q = - 36/3
⇒ q = - 12
Substitute q = - 2 in EQ(1)
⇒ - 12 + 6p = - 29
⇒ 6p = - 17
⇒ p = - 17/6
Therefore the equation equation can be written as
⇒ 6x⁴ - 29x³ + 40x² - 7x - 12 = 0
⇒ ( 6x² - 12x - 6 )( 6x² - 17x + 12 ) = 0
Solve the equations individually
6x² - 12x - 6 = 0
⇒ x² - 2x - 1 = 0
Use Quadratic formula
⇒x = ( 2 ± √( 4 + 4 ) )/2
⇒ x = ( 2 ± 2√2 )/2
⇒ x = [ 2( 1 ± √2 ) ] /2
⇒ x = 1 ± √2
Now, let's solve 6x² - 17x + 12 = 0
⇒6x² - 9x - 8x + 12 = 0
⇒3x( 2x - 3 ) - 4( 2x - 3 ) = 0
⇒( 3x - 4 )( 2x - 3 ) = 0
⇒ 3x - 4 = 0 or 2x - 3 = 0
⇒x = 4/3 or x = 3/2
Therefore the roots of equation are 1 ± √2, 4/3 and 3/2.
Theory of Equations
Answer:
1 ± √2 , 4/3, 3/2 are the roots
Step-by-step explanation:
Refer to attachment
