Question

how do i find the zeros of this quadratic equation by completing the square?
x2+10x+29

Answer 1

to understand how to complete a square, you must first realize what a complete square looks like

a complete square is of the form (x+n)^2; can you expand that out for me?

Answer 2

no i cant i dont think.whats the first step i could take in this problem?

Answer 3

(x+n)^2 is jsut (x+n) (x+n)

they go over some memory technique called "foil" if that rings a bell

Answer 4

yes i know how to do foil

Answer 5

then can you foil (x+n)(x+n) for me

Answer 6

Xsquared+xn+nx+Nsquared

Answer 7

good, lets clean that up into:

x^2 + 2n x + n^2 this is what a complete square looks like
lets compare that to what they gave you.


x^2 + 10 x + 29 notice that we would love to have an n^2 in this someplace
so lets find n

2n = 10, solve for n

Answer 8

n=5

Answer 9

great, then a complete square would look like:

x^2 + 10 x + 25

would you agree that we have:

x^2 + 10 x + 25 + 4 ??

Answer 10

where did the 4 come from?

Answer 11

well, we are given

x^2 + 10 x + 29, notice that 25+4 = 29

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since (x+n)^2 = x^2 + 2n x + n^2, we can compress the expanded parts back into its little package again

x^2 + 10 x + 25 + 4

(x^2 + 10 x + 25) + 4

(x+5)^2 + 4

and of course the zeroes or roots are when this equals zero

(x+5)^2 + 4 = 0 ; can you think of a way to solve for x?

Answer 12

we could have also just said: since 25-25 = 0, just add this zero to it

x^2 + 10 x + 29

x^2 + 10 x + 29 + 25 - 25

x^2 + 10 x + 25 + 29 - 25

x^2 + 10 x + 25 + 4

Answer 13

factor?

Answer 14

we cant really factor at the moment, but was have made life easier by compressing the complete square back into its package

Answer 15

(x+5)^2 + 4 = 0 ; -4

(x+5)^2 = -4 ; take the sqrt of each side to undo that ^2

x+5 = \(\pm \sqrt{-4}\) ; and finish off by -5

x = -5 \(\pm \sqrt{-4}\)

Answer 16

okay i see why you undid the ^2 so would that be my final answer?

Answer 17

it might need to be made more proper, something about an "i" ; but yes, thats the bulk of it

Answer 18

thanks i actually understand that haha

Answer 19

:) good luck

Answer 20

thank you:)

Answer 21

This is a great explanation of how completing the square works.
Thanks, @amistre64

Answer 22

\[x ^{2}+10x=0-29\]
\[adding both sides \left( \frac{ co-efficient of x }{2 } \right)^{2}i.e.,\left( \frac{ 10 }{ 2 } \right)^{2} or 25 ,we get\]
\[x ^{2}+10x+25=-29+25=-4=4\iota ^{2}\]
\[\left( x+5 \right)^{2}=\left( 2\iota \right)^{2}\]
\[x+5=\pm2\iota \]
solve and get the solution.