Question

Which of the following is a solution of x2 + 5x = -2?

Answer 1

i can help :D

Answer 2

it has to be a quadratic formula

Answer 3

solution i meant

Answer 4

no multiple choice?

Answer 5

kk

Answer 6

i have it

Answer 7

it'll be an image

Answer 8

k

Answer 9

first re-write the equation in 'standard' quadratic form:

\[ax ^{2}+bx+c=0\]

Answer 10

a=1 b=5 c=2

Answer 11

x^2-5x+2=0

Answer 12

ok - that's correct

Now this does not factorise. SO you need to solve by another method

Use the quadratic formula to get the two solutions to this equation:

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

Answer 13

im trying to do the formula

Answer 14

BTW - you mis typed the equation
x^2+5x+2=0

Answer 15

i got it -5+-sqroot 27 over 2

Answer 16

correct?

Answer 17

Yes correct BUT
just be careful: the WHOLE thing is /2

(-5+- sqr(27))/2

Answer 18

oh ok, next problem

Answer 19

hold on

b^-4ac = 5^2 -4(1*2) = 25-8 = 17

Answer 20

wasn't that i put down

Answer 21

so may i do next problem

Answer 22

Solve 3x2 + 4x = 2.

Answer 23

i don't understand the choices they have

Answer 24

um, nvm, i sorta had a blank mind and the answer was right there

Answer 25

1) Please post each new problem separately.
2) More important than understanding the answer choices is that you yourself learn how to solve these problems.
3) The first step towards solving 3x2 + 4x = 2 is to move that 2 to the left side of this equation. To accomplish that, subtract 2 from both sides.
doing this puts your equation into the required standard form: y=ax^2+bx+c=0

Answer 26

oh ok thx

Answer 27

then you have several choices of approach: quadratic formula, factoring, graphing.

Answer 28

i did the quadratic formula, and the answer i got was -2+- 2sqrt 10 over 3

Answer 29

is that correct?

Answer 30

Now please check your answer by substituting it back into the original equation.

Answer 31

how do i do that

Answer 32

I'm sure you've done this kind of check before. Supposing that you started with the equation 3x2 + 4x = 2 and that you came up with the possible answer x=1. You'd substitute x=1 into this equation:

3(1)^2+4(1)=2. Is this equation true or false?

Answer 33

false

Answer 34

7 doesn't equal 2

Answer 35

Right. That's not important; what is important is that you be able to check your possible solutions by substituting them back into the original equation.

Answer 36

'i did the quadratic formula, and the answer i got was -2+- 2sqrt 10 over 3'

Answer 37

What's your next step?

Answer 38

idk...

Answer 39

What did I just ask you to do, as an example?

Answer 40

just substitute

Answer 41

and i got it wrong

Answer 42

The whole point, Minias, is that if you want to check your possible roots, you'll need to substitute them back into the original equation. Have you tried that with your possible roots \[x=-2+- 2\sqrt 10 \over 3\] or
\[x=\frac{ -2\pm \sqrt{10} }{ 3 }?\]

Answer 43

Your 'x2 + 5x = -2' becomes \[1x^2+5x+2=0\]

Answer 44

oh

Answer 45

...and from this you can see that the coefficients 'a', 'b' and 'c' are 1, 5 and 2.

Answer 46

ahh i think i somewhat got it

Answer 47

I'm sorry, but I was focusing on the wrong problem. This is why I asked you to post each new problem separately.

Am I correct in believing that we're now focused on solving 3x2 + 4x = 2?

Answer 48

Please subtract 2 from both sides of this equation (assuming that you want to continue).

Answer 49

yes

Answer 50

3x2+4x-2

Answer 51

You've started with an equation and must end with an equation. Please type "=0" after your 3x^2+4x-2.

Answer 52

3x^2+4x-2=0

Answer 53

Good. Now, identify the coefficients a, b and c.

Answer 54

a=3 b=4 c=-2

Answer 55

Good. Now take those three coeff. values and substitute them into \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 56

how do i make that?

Answer 57

All I'm asking you to do is to substitute a=3, b=4 and c=-2 into the quadratic formula.

Answer 58

Hint: b=4, so we get\[x=\frac{ -4\pm \sqrt{4^2-4(a)(c)} }{ 2a }\]

Answer 59

substitute a=3 and b=-2 into this formula, please.

Answer 60

-4(3)(-2)

Answer 61

which would be +24

Answer 62

Yes, that's part of the discriminant, \[b^2-4ac,\], but you'll need to complete all of the required substitutions into \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 63

Again, if b=4, we get \[x=\frac{ -4\pm \sqrt{4^2-4(a)(c)} }{ 2a }\]

Answer 64

Please substitute a=3 and c=-2 into this formula.

Answer 65

4(3)(-2)
2(3)

Answer 66

Note: 4(3)(-2)=12(-2) = -24. Where did your 2(3) come from?

Answer 67

3 =a

Answer 68

I'm asking you to start with \[x=\frac{ -4\pm \sqrt{4^2-4(a)(c)} }{ 2a }\] and to substitute the values a=3 and b=-2. Your answer must look like\[x=\frac{ -4\pm \sqrt{4^2-4(a)(c)} }{ 2a }\] or it'd not be complete.

Answer 69

sry i meant -4

Answer 70

oh ok

Answer 71

\[\frac{-4\pm \sqrt{4^2-4(3)(-2)} }{ 2(3) }\]

Answer 72

like that?

Answer 73

Nice! and that reduces to\[x=\frac{-4\pm \sqrt{16+24} }{ 6 }\]

Answer 74

can you simplify this further?

Answer 75

\[\frac{ -4\pm \sqrt{40} }{ 6}\]

Answer 76

Very good. Be sure to label this:\[x=\frac{ -4\pm \sqrt{40} }{ 6}\]

Answer 77

and to reduce it. Note that Sqrt(40)=Sqrt(4)*Sqrt(10)=2Sqrt(10).

Answer 78

wt would 40 reduce to?

Answer 79

reduce it now, please:\[x=\frac{ -4\pm \sqrt{40} }{ 6}=\frac{ -4\pm2\sqrt{10} }{ 6 }=?\]

Answer 80

nvm

Answer 81

40 factors into 4 and 10, and 4 is a perfect square, so Sqrt(40)=2Sqrt(10).

Answer 82

\[\frac{ 4\pm \sqrt{10} }{ 3 }\]

Answer 83

-4 i meant

Answer 84

Please take another look at that:\[\frac{ -4\pm2\sqrt{10} }{ 6 }=\frac{ -4 }{ 6 }\pm \frac{ 2\sqrt{10} }{ 6 }=?\]

Answer 85

-4/6 reduces to -2/3.

2Sqrt(10)/6 reduces to what?

Answer 86

\[\frac{ -2\pm \sqrt{10} }{ 3}\]

Answer 87

Good. That's it. Any questions? Be sure to label your results.\[x=\frac{ -2\pm \sqrt{10} }{ 3}\]

Answer 88

one more

Answer 89

but i'll make another post on it

Answer 90

Please take a moment to think about what you've learned from solving these two problems. I'd suggest you take notes so that you have something to which to refer next time you encounter a problem requiring you to find the roots of a quadratic equation.

Answer 91

oh ok

Answer 92

I got all of them right, thx

Answer 93

Happy to hear that.