0

Guest

Courses New

find the solution of the equation 12x 4 -56x 3 +89x 2 -56x+12=0 i challenge you to find all the 4 real roots

find the solution of the equation 
12x4-56x3+89x2-56x+12=0
i challenge you to find all the 4 real roots

Grade:10

2 Answers

Arun
25750 Points
5 years ago
Note the symmetric coefficients. x=0 is not a root so start by dividing by x^2. This gives. 

12(x^2 + 1/x^2) - 56(x+1/x)+89 = 0 

Now subtract 12(x+1/x)^2 from both sides and simplify 

- 56(x+1/x)+65 = -12(x+1/x)^2 

Let u=x+1/x and you now have the quadratic equation 

12u^2 - 56u + 65 = 0 

(2u-5)(6u-13) = 0 

u=5/2 or u=13/6 

x+1/x=5/2 gives x=2 or x=1/2 and 

x+1/x = 13/6 gives x=3/2 or x=2/3.
 
 
ritika
21 Points
5 years ago
12x^4 – 56x^3 + 89x^2 – 56x + 12 = 0
Dividing by x^2;
12x^2 – 56x + 89 – 56/x + 12/x^2 = 0
(12x^2 + 12/x^2) – (56x + 56/x) + 89 = 0
12(x^2 + 1/x^2) – 56(x + 1/x) + 89 = 0
12[(x + 1/x)^2 – 2] – 56(x + 1/x) + 89 = 0
Let m = x + 1/x   …........(1)
12(m^2 – 2) – 56m + 89 = 0
12m^2 – 56m + 89 – 24 = 0
12m^2 – 56m + 65 = 0 
12m^2 – 30m – 26m + 65 = 0
6m(2m – 5) – 13(2m – 5) = 0
(2m – 5)(6m – 13) = 0
2m – 5 = 0 and 6m – 13 = 0
m = 5/2 and m = 13/6
 
From (1);
x + 1/x = 5/2 and x + 1/x = 13/6
2x^2 – 5x + 2 = 0 and 6x^2 – 13x + 6 = 0
2x^2 – 4x – x + 2 = 0 and 6x^2 – 9x – 4x + 6 = 0
(x – 2)(2x – 1) = 0 and (2x – 3)(3x – 2) = 0
x = 2, x = ½ and x = 3/2, x = 2/3

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More