solve the equation x^4+2x^3-4x^2-22x+40=0,given that the roots are in A.P
Answers
Given : x⁴ + 2x³ - 4x² - 22x + 40 = 0
To Find : Solve the equation
Solution:
Assume that roots are
a - 3d , a - d , a+d , a + 3d
Sum of roots = a - 3d + a - d + a + d + a + 3d = 4a
x⁴ + 2x³ - 4x² - 22x + 40 = 0
Sum of roots = - 2
=> 4a = - 2
=> a = -1/2
Product of roots = 40
=>( a - 3d) (a - d)(a+d)(a + 3d) = 40
=> (a² - 9d²)(a² - d²) = 40
=> (1/4 - 9d²)(1/4 - d²) = 40
=> (1 - 36d²)(1 - 4d²) = 640
4d² = y
=> (1 - 9y)(1 - y) = 640
=> 9y² - 10y + 1 = 640
=> 9y² - 10y - 639 = 0
=> 9y² - 81y + 71y - 639 = 0
=> 9y(y - 9) + 71(y - 9) = 0
=> (9y + 71)(y - 9) = 0
=> y = 9 y can not be - 71/9 as its 4d²
4d² = 9
=> d = ± 3/2
a = -1/2 , d = 3/2 or -3/2
roots are a - 3d , a - d , a+d , a + 3d
Roots are : - 5 , - 2 , 1 , 4
or 4 , 1 , - 2 , - 5
(x + 5)(x + 2)(x - 1)(x - 4) = x⁴ + 2x³ - 21x² - 22x + 40 = 0
So correct Question should be :
x⁴ + 2x³ - 21x² - 22x + 40 = 0
Roots are : - 5 , - 2 , 1 , 4
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