Question

Solve 2x^2+20x+3 =0 by completing the square

Answer 1

\[2x^2+20x+3 =0\]
\[2(x^2+10x)+3=0\]
\[2(x+5)^2-25(2)+3=0\]
\[2(x+5)^2-47=0\]
Solve for x

Answer 2

thanks, but why isn't 3 divided by two when you factorise?

Answer 3

There's a few techniques to solve this

Answer 4

which grade you in?

Answer 5

15 years old, not sure educational system in US but i think i'm in grade 10

Answer 6

My technique is,
\[ax^2+bx+c=0\]
Always make sure that "a" is 1
Factor out the coefficient of x^2
\[a(x^2+\frac{b}{a}x)+c=0\]
Then, divide 2 from b/a , then bring out b/2a (always minus sign) and square b/2a , then stick back "a" to b/2a
\[a(x+\frac{b}{2a})^2-(\frac{b}{2a})^2(a)+c=0\]

Answer 7

thank you, helped a lot with my revision :)

Answer 8

its quite confusing at first but if you get the hang of it, its the fastest method to solve a completing the quare