<span>Assume that,
x^2+4x-5 = 0 .......(1)
Then,
x^2+4x-5 = 0
x^2+5x-1x-5 =0
x(x+5)-1(x+5) = 0
(x+5) (x-1) = 0
We get x=-5 and x=1
Sub x=-5 in equ (1)
(-5)^2+4(5)-5 = 0
-25+20-5 = 0
-25+25= 0
0 = 0
Sub x=1 in equ (1)
(1)^2+4(1)-5 = 0
1+4-5 = 0
5-5 = 0
0 = 0
Therefore x value is -5 and 1</span>
Answer:
raising both sides of the equation to a certain power in order to eliminate radicals may result in the creation of extraneous roots
Step-by-step explanation:
Answer:
25.12
Step-by-step explanation:
Formula: 2πr, r = d/2
Given: π = 3.14, d = 8
Sub: r = 8/2
Simplify: r = 4
Sub: 2(3.14)(4)
Simplify: 6.28(4)
Solve: 25.12