Question

Repeated root of the equation 4x3-12x2-15x-4=0

Answer 1

Answer:

4x3-12x2+12x-4=0 

One solution was found :

                   x = 1

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(((4 • (x3)) - (22•3x2)) + 12x) - 4 = 0

Step  2  :

Equation at the end of step  2  :

((22x3 - (22•3x2)) + 12x) - 4 = 0

Step  3  :

Step  4  :

Pulling out like terms :

 4.1     Pull out like factors :

   4x3 - 12x2 + 12x - 4  = 

  4 • (x3 - 3x2 + 3x - 1) 

Checking for a perfect cube :

 4.2    x3 - 3x2 + 3x - 1  is not a perfect cube

Trying to factor by pulling out :

 4.3      Factoring:  x3 - 3x2 + 3x - 1 

Answer 2

Answer:

The repeated root of the given equation is 1.

Step-by-step explanation:

Given equation,

$4 {x}^{3} - 12 {x}^{2} - 15x - 4 = 0$

Let roots of this equation are a,a,b.

So, we can say

$a + a + b = \frac{12}{4} = 3$

$2a + b = 3.....(1)$

and

$a \times a \times b = \frac{4}{4} = 1 \\ {a}^{2} b = 1......(2)$

From equation (1) and (2),

$2 {a}^{2} + \frac{1}{ {a}^{2} } = 3$

$2 {a}^{3} - 3 {a}^{2} + 1 = 0$

$(a - 1)(2 {a }^{2} - a - 1) = 0$

$2 {a}^{2} - a - 1 = 0$

By solving this equation, we get

$a = \frac{1±3}{4} = \frac{4}{4} \: \: and \: \: \frac{ - 2}{4} = 1 \: \: and \: \: \frac{ - 1}{2}$

When a=1

$b = \frac{1}{ {1}^{2} } = 1$

and when

$a = - \frac{1}{2}$

b=4

Therefore repeated root of the given equation is 1.