Solve By Factoring

[nycmathguy]

Set 1.3
Question 33

See attachment.

Isn't it better to solve for x using the quadratic formula in terms of question 33?

 

[MathLover1]

x^2+(2sqrt(5))x+5=0....you can write 5 as sqrt(5)^2
x^2+(2sqrt(5))x+sqrt(5)^2=0............recognize square of the sum (a+b)^2=a^2+2ab+b^2

then you have

(x + sqrt(5))^2=0....factor
(x + sqrt(5)) (x + sqrt(5))=0

solution:
(x + sqrt(5))=0 => x =- sqrt(5))

 

[nycmathguy]

x^2+(2sqrt(5))x+5=0....you can write 5 as sqrt(5)^2
x^2+(2sqrt(5))x+sqrt(5)^2=0............recognize square of the sum (a+b)^2=a^2+2ab+b^2

then you have

(x + sqrt(5))^2=0....factor
(x + sqrt(5)) (x + sqrt(5))=0

solution:
(x + sqrt(5))=0 => x =- sqrt(5))

I will post a few more like this one to get practice.
Where did (x + sqrt{5}) come from?