Question

Solve 5x2 - 7x + 2 = 0 by completing the square.



What is the constant added on to form the perfect square trinomial?


49/4
49/25
49/100

Answer 1

\[5x^{2} - 7x +2 = 0\]\[5(x^{2} - \frac{7}{5}x) + 2 = 0\]\[5(x^{2} - \frac{7}{5}x + \frac{49}{100} - \frac{49}{100}) + 2 = 0\]\[(x-\frac{7}{10})^{2} - \frac{49}{20} + 2 = 0\]\[(x-\frac{7}{10})^{2} - \frac{9}{20} = 0\]

Answer 2

Really good job.

Answer 3

what would i do after your last step?

Answer 4

@Jhannybean thank you! :)

Answer 5

That is the final step. otherwise youd be solving for x and that involves a multitude of other steps.

Answer 6

yep!

Answer 7

no its says to solve for "x" using completing the square

-you want to isolate the squared term, then take sqre root

Answer 8

oh you DO want to solve haha ah......

Answer 9

oh yes, sorry i read that wrong! thanks for correcting it @dumbcow

Answer 10

im confused

Answer 11

\[(x-\frac{7}{10})^{2} = \frac{9}{20}\]\[x = \sqrt{\frac{9}{20}}+\frac{7}{10}\]

Answer 12

\[(x-\frac{7}{10})^{2} - \frac{9}{20} = 0\]\[(x-\frac{7}{10})^2 = \frac{9}{20}\]sqare root both sides\[\sqrt{(x-\frac{7}{10})^2}=\sqrt{\frac{9}{20}} \]\[x-\frac{7}{10} = \pm \frac{3}{\sqrt{20}}\]\[x= \frac{7}{10} \pm \frac{3}{20}\]

Answer 13

\[ x= \frac{7}{10} \pm \frac{3}{2\sqrt{5}}\]

Answer 14

i forgot the square root on my second to last post.

Answer 15

which answer would i pick then ?

Answer 16

here are some general formulas to use:
given a quadratic equation
\[ax^{2} +bx +c = 0\]
completing the square yields
\[a(x+\frac{b}{2a})^{2} +(c-\frac{b^{2}}{4a}) = 0\]
solving gives the quadratic formula
\[x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

Answer 17

the "constant" added to coplete the square would be the value you'd obtain by finding the value "c". You have to fit your equaion into the quadratic form \[\large ax^2 +bx+c=0\] and to find your value of c, you evaluate your value of "b" :-

\[\large c : \large \left(\frac{-7/5}{2}\right)^2 = \frac{49}{100}\]

Answer 18

i thought that i was supposed multiply the 5 to the 2 (c term) and then subtract that number to both sides. then after that dont i take half of the negative seven and then put it into x+3.5

Answer 19

thank you @Jhannybean

Answer 20

Yeah...what you're talking about can also be solved like this.you don't necessarily HAVE to complete the square. I'll show you a different method.

Answer 21

\[\large 5x^2 -7x+2=0\]\[\large \color{green}5x^2-7x+\color{green}2=0\]\[\large x^2 - 7x + 10 =0\]\[\large (x-5)(x-2)=0\]\[\large (x-\frac55)(x-\frac25)=0\] Simplify whatever you can. \[\large (x-1)(5x-2)=0 \] Solve for x's. \\[\large x-1 =0 \]\[\large 5x-2 =0 \]\[\large x = 1 \ ,\ \frac25\]

Answer 22

Gotta go! -Jhannybean takes leave-

Answer 23

isnt that finding the factors though?

Answer 24

Mmhmm. That is. it's just another method. :D

Answer 25

ah ok thank you :D

Answer 26

np :D