OK can someone please check my answer for me the problem is x^2-2x-1=0 it states that it has 2 solutions it wants me to list the smallest solution which i calculated to me 0 but the software marks that wrong is there something i did wrong?

Question

jim_thompson5910 · Answer

What 2 solutions did you get?

anonymous · Answer

um i believe I got 0 and i forgot what the other was i don't have the work all together right now but it was bigger then 0

anonymous · Answer

i know that for sure

jim_thompson5910 · Answer

0 isn't a solution


x^2-2x-1=0

(0)^2-2(0)-1=0 ... plug in x = 0

0 - 2(0) - 1 = 0

0 - 0 - 1 = 0

-1 = 0 ... this is false, so x = 0 isn't a solution

anonymous · Answer

ok i see my error but can you help me solve the problem then because i don't know what I'm really doing

jim_thompson5910 · Answer

You have to use the quadratic formula to solve for x.

jim_thompson5910 · Answer

Do you remember that formula?

anonymous · Answer

yes i remember it

anonymous · Answer

thats what I'm having the issue with someone helped me get the numbers i need to plug in but its still confusing me

javk · Answer

It would be difficult to solve using factorization so use the formula 
$$x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }$$

anonymous · Answer

ok thats what I used originally I was helped to get the numbers to plug in for the variables we got -2 for b 1 for a -1 for c

anonymous · Answer

but somewhere during the solving process it went wrong i don't really know what i did wrong

jim_thompson5910 · Answer

what is $\Large b ^{2}-4ac$ equal to?

anonymous · Answer

that would be -2^2 which would be -4 and -4(1)(-1) would be-4+4 which would be 0 right?

jim_thompson5910 · Answer

b^2 is not -2^2 = -4

b^2 = (-2)^2 = 4

anonymous · Answer

ok sorry your right i don't know what i was thinking its getting late i just want to hurry and finish this last problem

jim_thompson5910 · Answer

so what is b^2 - 4ac equal to?

anonymous · Answer

ok so b^2 = 4 and -4(1)(-1)= 4 so its 4+4 which equals 8 right?

jim_thompson5910 · Answer

correct

jim_thompson5910 · Answer

$$\large x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }$$

$$\large x=\frac{ -(-2) \pm \sqrt{(-2) ^{2}-4(1)(-1)} }{ 2(1) }$$

$$\large x=\frac{ 2 \pm \sqrt{8} }{ 2 }$$

how do we simplify sqrt(8) ?

anonymous · Answer

it can be broken down to 2*2*2 and you can take 2 2s so it would then be 2 square root 2 right?

jim_thompson5910 · Answer

yes

$$\large x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }$$

$$\large x=\frac{ -(-2) \pm \sqrt{(-2) ^{2}-4(1)(-1)} }{ 2(1) }$$

$$\large x=\frac{ 2 \pm \sqrt{8} }{ 2 }$$

$$\large x=\frac{ 2 \pm 2*\sqrt{2} }{ 2 }$$

jim_thompson5910 · Answer

then we can factor out 2 from the numerator and cancel the '2's

$$\large x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }$$

$$\large x=\frac{ -(-2) \pm \sqrt{(-2) ^{2}-4(1)(-1)} }{ 2(1) }$$

$$\large x=\frac{ 2 \pm \sqrt{8} }{ 2 }$$

$$\large x=\frac{ 2 \pm 2*\sqrt{2} }{ 2 }$$

$$\large x=\frac{ 2(1 \pm \sqrt{2}) }{ 2 }$$

$$\large x=\frac{ \cancel{2}(1 \pm \sqrt{2}) }{ \cancel{2} }$$

$$\large x=1 \pm \sqrt{2}$$

$$\large x=1 + \sqrt{2} \ \ \text{or} \ \ x = 1-\sqrt{2}$$

anonymous · Answer

ok so x= 1-square root of 2 would be my answer since its asking for the smallest solution?

jim_thompson5910 · Answer

correct

anonymous · Answer

ok thanks for your help

anonymous · Answer

now to go sleep lol so i can start thinking straight again