1. anonymous

Start off by putting into $ax^2+bx+c=0$ form

ok, wait let me solve it first

3. anonymous

ok, x^2 +1x -7

5. anonymous

Good now we have the quadratic formula which is the following $x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$ where a = 1, b= 1, and c = -7

ok, wait again sorry

8. anonymous

Are you sure?

yes

no no i am wrong wait

the answer is 2.15 and -3.15

i took the wrong equation

13. anonymous

$x= \frac{ -1 \pm \sqrt{1^2-4(1)(-7)} }{ 2(1) } \implies x = \frac{ -1 \pm \sqrt{29} }{ 2 }$ correct?

yes

15. anonymous

That should $x \approx -3.19,~~ \text{and}~~ x \approx 2.19$

can i ask you an another quetion pls

17. anonymous

Sure thing

Use Quadratic formula to solve for x: $x + \frac{ 1 }{ x+2 } = 4$

19. anonymous

Same thing as earlier, try putting it into $ax^2+bx+c=0$

how will i do that

21. anonymous

You can multiply through by (x+2)

??

23. anonymous

I mean you can find a common denominator which in this case is (x+2) if you like

can we cross multiply

25. anonymous

You can do as following $\frac{ 1 }{ x+2 }=4-x$ now you may do as you like

26. anonymous

$\frac{ 1 }{ (x+2) } = (4-x)$

ok

wait

is it = to 4x-x-5??

30. anonymous

Hmm, not quite, $\frac{ 1 }{ (x+2) } = (4-x) \implies 1 = (4-x)(x+2)$

yeah then we multiply (4-x)(x+2) = 4x - x +6

an we bring it to the other side

-4x + x -6 + 1

= -4x + x -5 = 0

right??

36. anonymous

Remember, we want our equation in $ax^2+bx+c=0$ form, so what you wrote, does that make sense? Think of what you did in the previous problem

x^2-4x-5

= 0

39. anonymous

$(4-x)(x+2)=1 \implies 4x+8-x^2-2x =1 \implies 2x+8-x^2=1$ so we can set this up as $x^2-2x-7=0$

ok. sorry i was offline, i had a lag

answer is = 1.9, -0.9 right??

@iambatman

43. anonymous

Can you show how you got that?

ok |dw:1444558479751:dw|

|dw:1444558581896:dw|

46. anonymous

What is a, b, and c?

b =-2, c -7

48. anonymous

$x = \frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-7)} }{ 2(1) } = \frac{ 2 \pm \sqrt{32} }{ 2 }$

49. anonymous

a = 1, b = -2, c = -7