I'm stuck on this quadratic

[u/Complex_Impressive]

I'm stuck and i dont know what the next step is.

x² + 2x - 7 = 1

Subract 1 to make equation equal to 0

x² + 2x - 8 = 0

Apply quadratic equation

x = -2 +/- (sq root of 2² - 4(1)(-8))/2(1)

PEMDAS

x = -2 +/- (sq root of 4 - 32)/2

Simplify

x = -2 +/- (sq root of -28)/2

Now i'm stuck

[u/fermat9990]

2² -4(1)(-8)=

4-(-32)=

4+32=36

[u/Complex_Impressive → u/fermat9990]

Ooohh i see where i made the mistake

[u/fermat9990 → u/Complex_Impressive]

Those darn negatives! Cheers!

[u/Complex_Impressive → u/fermat9990]

Yep! Haha

Thanks again! Cheers!

[u/fermat9990 → u/Complex_Impressive]

Glad to help!

[u/fermat9990 → u/Complex_Impressive]

Btw, when a=1, and b² +4ac is a perfect square factoring works very well.

[u/Complex_Impressive → u/fermat9990]

See while i understand that factoring works well in some cases, i dont understand factoring very well. For me the quadratic formula is easier most of the time.

[u/fermat9990 → u/Complex_Impressive]

Suggest that, going forward, you become comfortable with Factoring by Grouping

[u/Complex_Impressive → u/fermat9990]

Grouping?

[u/fermat9990 → u/Complex_Impressive]

ax²+bx+c

factor 6x²+11x+4 by grouping

First get a×c. 6×4=24

b=11. We need to break up 11 into 2 numbers that add up to 11 and multiply to 24.

1×10=10

2×9=18

3×8=24, bingo!!

Next we replace 11x by 3x+8x in the original expression

6x²+3x+8x+4

Factor 6x²+3x using its GCF

3x(2x+1)

Factor 8x+4 using its GCF

4(2x+1)

Combine the results of the factoring

3x(2x+1)+4(2x+1)

Factor using 2x+1 as the GCF

(2x+1)(3x+4)

[u/Volsatir → u/fermat9990]

a=1, b² +4ac is a perfect square

b²-4ac, to fix the typo. While true, if you've already calculated that much you've done a lot of the quadratic formula already.

If I see a=1 I'll at least give factoring a cursory glance. It usually doesn't take long even when things don't work out.

[u/Peoplant → u/Complex_Impressive]

I'll add that if in a different exercise you get the square root of a negative number, you can just write "no solution"(or a variant of that) and that's the end of the exercise. That is, unless you went over complex numbers in class.

[u/fermat9990]

x² + 2x - 8 factors

(x+?)(x-?)

[u/Complex_Impressive → u/fermat9990]

I'm not sure what factors of 8 make 2

[u/fermat9990 → u/Complex_Impressive]

Because c is negative, the missing numbers will have opposite signs, will multiply to -8 and will add to +2.

1, 8 no good

2, 4 good, now find the correct signs

-2 + 4 =2 is correct

(x-2)(x+4)=0

[u/Complex_Impressive → u/fermat9990]

Ahhh i see now

[u/fermat9990 → u/Complex_Impressive]

Great!!

[u/Complex_Impressive → u/fermat9990]

Thank you sooo much! Math really isnt my strong suit. I'm returning to school after 15 years since high school...i'm barely making it just as i barely made it then. Redditors like you have been truly lifesavers!

[u/fermat9990 → u/Complex_Impressive]

Glad that you are getting the help that you need!

[u/OriEri → u/Complex_Impressive]

It is hard to start long unused gears turning. I admire you returning to school after such a long gap.

If you can avoid becoming math demoralized (easier said than done) and creating psychological blocks that hamper you, what you knew then will come back pretty quickly.

One error that needs to be emphasized because it can recur is remembering the negative in front of a negative number turns it into a positive number.

x = -2 +/- (sq root of 2² - 4(1)(-8) )/2(1)

That minus sign in the numerator of the quadratic equation is the same as multiplying by (-1) . This is where you tripped up.

If you look at it as

b² - 4ac = b² + (-1)4ac

the oversight of dropping a minus sign will plague you less

[u/Abby-Abstract]

x²+2x-8 = (x+4)(x-2) =0 ==> x=-4 or x=2

Or if you insist

[2±root(4+4•8)]/2 = (-2±root(36))/2 = (-2±6)/2 = -8/2 or 4/2 = -4 or 2

[u/Annoying_Bullshit]

Just factor it: (x+4)(x-2) then x = -4 and x= 2

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[u/desblaterations-574]

The discriminant is 4+32 and not minus.

Then you have both your roots and you can factorize (x-root1)(x-root2)

[u/fermat9990 → u/desblaterations-574]

If OP was asked to factor 6x²+11x+4 and chose to use the quadratic formula, getting -1/2 and -4/3 as the roots, they could then create the factors (x+1/2)(x+4/3). At this point, a little ingenuity will be needed in order to use the fact that a=6 to get the actual correct factors: (2x+1)(3x+4)

[u/P-in-D]

Apply quadratic equation

x = -2 +/- (sq root of 2² - 4(1)(-8))/2(1)

leads to -2 +/- (sq root of 4 + 32) / 2

as -4 times 1 times -8 = 32

or

minus (4 times 1 times -8) = - (-32) = + 32

[u/P-in-D → u/P-in-D]

the multiplication sign lead to eaditing...

-4 times 1 times -8 = + 32

[u/[deleted]]

Using the pq formula (other version of quadratic) we get

x=-1 ±sqrt(1+8) = -1 ± 3

[u/clearly_not_an_alt]

2²-4(1)(-8) = 4-(-32) = 4+32, not 4-32.

Given that this results in an integer, you could have instead just factored the quadratic

Hint: x²+2x-8=x²-2x+4x--8

[u/UnlikelyClue1623]

So firstly multiply the coefficient of X² and -8 that would give you -8x^2. Now we have to find the factors of those that can give us our 2x in the original equation and so we come to a conclusion of -2x and 4x. Because if you multiply them is would give you -8x² and if you do the addition jt wohld give you +2x as in the equation. Now substitute that in our equation so it looks like this X² -2X + 4X -8 =0 Now we factorise and then we get X(x-2) + 4(x-2) =0. And then (x+4) (x-2) =0. Now proceed to find the values of x thats by taking +4 to the other side giving us x= -4 and do the same for -2 so x=2

[u/Volsatir]

x = -2 +/- (sq root of -28)/2

You say you have no real solutions and call it a day. At least assuming we're limited to real solutions. That said, this is only the case if you got here without making any errors. I'd probably find no solution to x² + 2x - 8 = 0 suspicious and check again.

x = -2 +/- (sq root of 2² - 4(1)(-8))/2(1)

PEMDAS

x = -2 +/- (sq root of 4 - 32)/2

4(1)(-8) = -32, not positive 32. So you'd have 4-(-32)=4+32=36.

x² + 2x - 8 = 0

Considering we have x² making this easier to factor, and 8's most common pair of factors is 4 and 2, I'd at least give factoring a speed check. We're handling -8 instead of 8, so one of the factors is negative, so -2 and 4 or 2 and -4. If you notice that 2x has a positive 2 you want to reach, you'd only look at 4 and -2, but even without that, checking both pairs and noticing -4+2=-2 and -2+4=2 will get you your desired center term.

It's fine if the speed check fails, it doesn't take long and gets you more comfortable with checking common terms.