Question

QUADRATIC EQUATION: x^2 + 4x − 4 = 0

Could someone please show me how to solve a question such as this? Multiple methods would be helpful as well. ☻

edit: I'm aware the distance formula might work for a quadratic like this, however, I've always been really poor at executing it...

Answer 1

Okay.
You know. Every quadratic equation's heart is it's discriminant. A general quadratic equation is ax^2 + bx + c. The expression b^2 - 2ac is called its discriminant.

If the discriminant is greater than zero, the equation has two different real roots. Or solutions. These actually exist. If it is less than zero. These solutions donot exist, or lets say, are imaginary.

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

is the formula to directly get solutions of an equation.

Just substitute the values of a,b,c ( which are 1, 4 , -4 ). This is the best way only because it has sort of irrational roots.

Answer 2

"The expression b^2 - 4ac is called its discriminant."

Just to add more on discriminant :
If it is a perfect square, then the original quadratic can be factored easily.
So, the factoring method becomes the best in those cases.
(In this case, it isn't, but good to know.)

If it is negative, then the quadratic cannot be factored, and their roots are imaginary.

If it is just a positive which is not a perfect square,
then the roots will be rational, and either
1. Completing the square OR
2. Using the quadratic formula
is the method that can be used to find the roots.

Answer 3

Both of your guys' explanations were really helpful. I had forgotten all about the whole irrational solutions and knowledge of the discriminant.

I think where I struggle with most in terms of solving quadratics like these is when it can't be factored so easily, or the square root of one of the numbers ends up being imaginary. For example, this is how far I've gotten with this problem (by means of the distance formula) and where I'm stuck at:

\[x = \frac{ -4\pm \sqrt{4^{2}-4(1)(-4)} }{ 2(1) }\]

\[x = \frac{ -4\pm \sqrt{16-4(-4)} }{ 2(1) }\]

\[x = \frac{ -4\pm \sqrt{16+16} }{ 2(1) }\]

\[x = \frac{ -4\pm \sqrt{32} }{ 2(1) }\]

From this point, I'm not sure if I divide everything by 2, or if I'm supposed to take care of the \[\sqrt{32}\] . Am I right up to this point, or am I missing anything? Also, what should I do once I reach this point?

Answer 4

It seems to be the end parts that I struggle with... that whole order of operations, and what goes where afterwards, if that makes any sense. Thank you guys so much for sticking with this question and answering, by the way! I'm taking an online Algebra class and I'm really struggling with it, as it isn't very explanatory.

Answer 5

@hartnn @AakashTomar

Answer 6

you're right.
You can now distribute the denominator's 2 to numerators.

also,
\(\sqrt 32\) can be simplified

Answer 7

Okay.
\[\sqrt{32} = 4\sqrt{2}\]
Now cancel 2 from numerator and denominator both. One root is the expression with the plus sign and one is with the minus sign.
x= Expression 1 and x= Expression 2.

Answer 8

Is this what you guys are referring to? Still a little shaky at this point. Also, I feel like this should be common knowledge, but how did you simply \[\sqrt{32}\] to \[4\sqrt{2}\] ? Is this something that's kinda more of a mental understanding, or does it require a calculator? (As you can tell, my brain is NOT wired for mathematics.)

\[x = \frac{ -4\pm4\sqrt{2} }{ 2 }\]
\[x = -2 \pm 2\sqrt{2}\]

Answer 9

\(\sqrt{32} = \sqrt{2\times 2 \times 2 \times 2\times 2} = \sqrt {2^2 \times 2^2 \times 2} = \sqrt2^2 \times \sqrt 2^2 \times \sqrt 2 = (2\times 2 )\sqrt 2\)

see if that makes sense?

Answer 10

\(x = -2 \pm 2\sqrt{2}\)
is absolutely correct :)

Answer 11

That does make a little more sense, thank you!
So out of curiosity then, when dividing the entire equation by two, why doesn't the square root of 2 become 1 or another number? Why does it just stay the same?

Answer 12

Also, how can I award you lovely individuals on this site for helping me out? ☻ You've put a lot of effort into helping me understand, and I'm especially grateful!

Answer 13

1. you mean why \(4\sqrt2/2 = 2\sqrt 2\)
and not something else?

2. you can click on the "Best Response" Tab on any one of our answers. :)

Answer 14

1. Yes, exactly. I guess in my brain, when I see that 2 in the denominator, I want to divide EVERYTHING by 2, including the square root of 2. I guess I'm just curious as to why it stays the same.
2. I will definitely do that, thank you. ☻

Answer 15

1. Thats how multiplication and division work.

say 10*6/2, (10*6/2 = 60/2 = 30)
thats not 10/2 * 6/2. (10/2 * 6/2 = 5*3 = 15)

but (10+6)/2 = 10/2 + 6/2.

You can separate out the numerator terms only if there is + or - sign in between.

Answer 16

Oh, duh. Wow. Common knowledge.

Thank you so much Hartnn! As well as you, @AakashTomar! So greatly appreciated! ♥

Answer 17

2. welcome ^_^