Solve x^{3}+5x^{2}=4x+20 algebraically.

tripiverded9

tripiverded9

Answered question

2021-12-29

Solve x3+5x2=4x+20 algebraically.

Answer & Explanation

godsrvnt0706

godsrvnt0706

Beginner2021-12-30Added 31 answers

Solution:
Given: x3+5x2=4x+20
x2(x+5)=4(x+5)
x2(x+5)4(x+5)=0
(x+5)(x24)=0
(x+5)(x2(2)2)=0
(x+5)(x+2)(x2)=0
x=2,2,5
Hence value of x=2,2,5
Steve Hirano

Steve Hirano

Beginner2021-12-31Added 34 answers

We have to solve the above equation algebraically
x3+5x24x20=20=0
x2(x+5)4(x+5)=0
(x24)(x+5)=0
(x+2)(x2)(x+5)=0
{a2b2=(a+b)(ab)}
(x+2)=0 or (x2)=0 or (x+5)=0
x=2,x=2,x=5
Hence, x=2,2,and5 are the solution of above equation.
karton

karton

Expert2022-01-04Added 613 answers

x3+5x2=4x+20х2(x+5)=4(x+5)х2(x+5)4(x+5)=0(x+5)(х24)=0x+5=0 or x24=0x1=5 or x2=2x3=2Answer: х=5,2,2

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