Equation help?

Question

I need help with these equations !!!????

How do I solve these problems

(1) x^2 + 5 = 21


(2) 5(x-2)^2 = 3


(3) x^2 + 8x + b ( I have to find the constant term "b" that should

be added to make this equation a perfect sqaure trinomial.

Thanks in advance !

Answer 1

1. X^2 = 21-5
X^2 = 16
X = 4 and -4

2. 5(x-2)^2 = 3
5(x-2)(x-2) = 3
5(x^2-4x+4) = 3
5X^2-20x+20 = 3
5X^2-20x+17 = 0

Use quad. formula:

x = -(-20) =/- sqrt (400-340) / 17
x = (20 +/- sqrt 60) / 17
x = (20 +/- 2(sqrt 15)) / 17

3. X^2 + 8x + b

(X+4) (X+4) = X^2+8x + 16

so, b =16

Answer 2

(1) x^2=21-5 = 16 so x=+-4

(2)(x-2)^2 = 3/5 so x-2 = +- sqrt(3/5) and x = 2 +- sqrt(3/5)

(3) put it as (x+c)^2 = x^2 +2cx +c^2 so 2c= 8 c=4 and b=c^2 so b=16 and you have (x+4)^2 x^2+8x+16

Answer 3

x^2 + 5 = 21 take away 5 from both sides of the equals sign to give
x^2 = 16
square root both sides to give
x = 4 or x = -4 (there are two answers as there are two square roots of 16)