Question

How do you solve a perfect square trinomial if it has a number in front of the first x value?
For example: 25x^2+10x+1=9

Answer 1

I know how to solve it if it was just x^2+10x+1=9, but i dont know what to do with the first 25

Answer 2

5x^2+10x+1=9
5x^2+10x +1 - 9 = 0
5x^2+10x - 8 =0

you'll get a = 5, b = 10, and c = -8
substitute all of them to quadratic formula

\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

Answer 3

thank you

Answer 4

If I put it into 5x^2+10x - 8 =0 that form wouldn't that be considered completing the square and i could just put all the x values on one side and divide each side by 25? then i could put it into b/2^2 to solve for x

Answer 5

also how did you get the 5x^2 instead of the 25? @gerryliyana

Answer 6

sorry wrong typo!

Answer 7

i mean 25

Answer 8

25x^2+10x+1=9
25x^2+10x +1 - 9 = 0
25x^2+10x - 8 =0

you'll get a = 25, b = 10, and c = -8
substitute all of them to quadratic formula

Answer 9

\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]
\[x = \frac{ -(10) \pm \sqrt{10^2 - 4*25*(-8)}}{ 2*25 } = \frac{ -10 \pm \sqrt{100 - (-800)} }{ 50 } \]
\[x=\frac{ -10 \pm \sqrt{900} }{ 50 } = \frac{ -10 \pm 30 }{ 25 }\]

Answer 10

So, you have

\[x = \frac{ -10 + 30 }{ 50 } = \frac{ 20 }{ 50 } =\frac{ 2 }{ 5 }\]
and
\[x=\frac{ -10-30 }{ 50 } = \frac{ -40 }{ 50 } = -\frac{ 4 }{ 5 }\]

Answer 11

oh , I understand now thank you!!

Answer 12

you are very wlcome :)