Question

(Please Help I have Quiz tomorrow)
How would you solve 4x^2-49=0
[without quadratic formula]

Answer 1

You know what PEMDAS is correct?

Answer 2

yes

Answer 3

\[4x^2-49=0\]

Answer 4

Alright then, first off lets start with whats 0+49

Answer 5

49

Answer 6

Let change the equation to 4x^2=49

Answer 7

whats 49/4?

Answer 8

\(\bf 4x^2-49=0\implies 4x^2=49\implies x^2=\cfrac{49}{4}\quad taking\quad \sqrt{\qquad }
\\ \quad \\
\sqrt{x^2}=\pm\sqrt{\cfrac{49}{4}}\implies x=\pm\cfrac{\sqrt{49}}{\sqrt{4}}\)

Answer 9

12.25. Now we need to fine X^2=12.25

Answer 10

3.5. x=3.5

Answer 11

Did I explain it well enough?

Answer 12

sorry I am having technical difficulties.

Answer 13

wait i am looking over your solution.

Answer 14

Now to check the work. 4*3.5^2-49=0 3.5*3.5=12.25 12.25*4=49 49-49=0

Answer 15

where did the 3.5 come from?

Answer 16

To be honest, I was being lazy so I did this= http://www.bing.com/search?q=x%5e2%3d12.25&qs=HS&pq=x%5e2&sc=8-3&sp=1&FORM=QBRE&cvid=709734e95e1449b8b20958bf9a237e36

Answer 17

ins't \[x^2 \] basically x*x

Answer 18

Yes ma'am, it's a exponent

Answer 19

so if it said x^3 it would be x*x*x

Answer 20

oh well I will check on it thanks

Answer 21

\(\bf 4x^2-49=0\implies 4x^2=49\implies x^2=\cfrac{49}{4}\quad taking\quad \sqrt{\qquad }
\\ \quad \\
\sqrt{x^2}=\pm\sqrt{\cfrac{49}{4}}\implies x=\pm\cfrac{\sqrt{49}}{\sqrt{4}}\qquad {\color{blue}{ 49=7^2\qquad 4=2^2}}\quad thus
\\ \quad \\
x=\pm\cfrac{\sqrt{49}}{\sqrt{4}}\implies x=\pm\cfrac{\sqrt{7^2}}{\sqrt{2^2}}\implies x=\pm\cfrac{7}{2}\)

Answer 22

I kinda said that... but I guess yours makes a bit more since since you are showing her the equations

Answer 23

there's an easier way: notice that \[4x^2 = (2x)^2\]and \[-49 = -(7)^2\]We can factor this as a difference of squares:
\[a^2-b^2 = (a+b)(a-b)\]
\[(2x-7)(2x+7) = 0\]\[2x-7=0\]\[2x=7\]\[x=\frac{7}{2}\]\[2x+7=0\]\[2x=-7\]\[x=-\frac{7}{2}\]

so the solutions are \[x = \pm\frac{7}{2}\]

Answer 24

Thx for the medal btw

Answer 25

Thank you mr.whpalmer This is how i am doing these polynomials.

Answer 26

Oh, I forgot to add (I got distracted) do solve equations like these, you just reverse PEMDAS if you didn't catch on already

Answer 27

ok