Verified answer

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