Equations in Quadratic Form: Solving 3x^2 - 2x - 2 = 0

[KH%3]

Hello there

So I am studying for a Test, and came over this issue with one of my practice equations. I more often than not, struggle a bit with the Quadratics when it comes down to order, basicly doing something in a different order, than my solution does.
On this one however, I can not figure out, why I have to obey, and can't do it along my own way (even if it may be a bit more complicated). As said, this usually works out, after a while of trouble shooting.

This is the URL to my Question/Solutions on Symbolab (I'm not sure if you can access it, without the premium Membership)

And following i'll show you my try:


(red marks are what i forgot in the first run, then corrected it)
You can see the GREEN 7/9 on the right side, this is where Symbolab starts doing something different than me. They combine them both, to one single term. However not doing this should stil get me to the correct solutions, in my knowledge.

On the bottom, red is my solution(s), IGNORE the green ones.
I think the Solutions should be: x=1.215,x=0.549
However Symbolab only pleases me with:


I'd be very glad if someone can explain to me what I am doing wrong
So I continue, and try solving it.

I'd be glad if someone can explain my mistake, and calm my mind. I really can't imagine symbolab bugging out on me and giving me a wrong answer.

Thanks for the help in advance
KHper3

 

[KH%3]

Equations in Quadratic Form

Hello there

So I am studying for a Test, and came over this issue with one of my practice equations. I more often than not, struggle a bit with the Quadratics when it comes down to order, basicly doing something in a different order, than my solution does.
On this one however, I can not figure out, why I have to obey, and can't do it along my own way (even if it may be a bit more complicated). As said, this usually works out, after a while of trouble shooting.

This is the URL to my Question/Solutions on Symbolab (I'm not sure if you can access it, without the premium Membership)

And following i'll show you my try:

View attachment 10377
(red marks are what i forgot in the first run, then corrected it)
You can see the GREEN 7/9 on the right side, this is where Symbolab starts doing something different than me. They combine them both, to one single term. However not doing this should stil get me to the correct solutions, in my knowledge.

On the bottom, red is my solution(s), IGNORE the green ones.
I think the Solutions should be: x=1.215,x=0.549
However Symbolab only pleases me with:
View attachment 10378

I'd be very glad if someone can explain to me what I am doing wrong
So I continue, and try solving it.

I'd be glad if someone can explain my mistake, and calm my mind. I really can't imagine symbolab bugging out on me and giving me a wrong answer.

Thanks for the help in advance
KHper3

 

[Dr.Peterson]

So I am studying for a Test, and came over this issue with one of my practice equations. I more often than not, struggle a bit with the Quadratics when it comes down to order, basicly doing something in a different order, than my solution does.
On this one however, I can not figure out, why I have to obey, and can't do it along my own way (even if it may be a bit more complicated). As said, this usually works out, after a while of trouble shooting.
You can see the GREEN 7/9 on the right side, this is where Symbolab starts doing something different than me. They combine them both, to one single term. However not doing this should stil get me to the correct solutions, in my knowledge.

On the bottom, red is my solution(s), IGNORE the green ones.
I think the Solutions should be: x=1.215,x=0.549
However Symbolab only pleases me with:
View attachment 10378

I'd be very glad if someone can explain to me what I am doing wrong
So I continue, and try solving it.

I'd be glad if someone can explain my mistake, and calm my mind. I really can't imagine symbolab bugging out on me and giving me a wrong answer.

You did fine until you had the expression \(\displaystyle \frac{2}{3}+\left(\frac{1}{3}\right)^2\), and claimed to take its square root by taking the square root of each term separately. That is wrong!

There is no rule that \(\displaystyle \sqrt{a+b} = \sqrt{a} + \sqrt{b}\). In fact, you can see that it is wrong from an example: \(\displaystyle \sqrt{4+9} = \sqrt{13}\), but \(\displaystyle \sqrt{4} + \sqrt{9}=2+3=5\).

You need to add the fractions together and take the square root of the result, simply because that is what the expression says. There's nothing wrong with simplifying earlier or later than someone else does, but you can't "simplify" by doing something that changes the value of an expression.

The order in which you do things isn't important. It is whether you do things that are correct. Never do anything without knowing it is valid.

 

[JeffM]

Hello there

So I am studying for a Test, and came over this issue with one of my practice equations. I more often than not, struggle a bit with the Quadratics when it comes down to order, basicly doing something in a different order, than my solution does.
On this one however, I can not figure out, why I have to obey, and can't do it along my own way (even if it may be a bit more complicated). As said, this usually works out, after a while of trouble shooting.

This is the URL to my Question/Solutions on Symbolab (I'm not sure if you can access it, without the premium Membership)

And following i'll show you my try:

View attachment 10377
(red marks are what i forgot in the first run, then corrected it)
You can see the GREEN 7/9 on the right side, this is where Symbolab starts doing something different than me. They combine them both, to one single term. However not doing this should stil get me to the correct solutions, in my knowledge.

On the bottom, red is my solution(s), IGNORE the green ones.
I think the Solutions should be: x=1.215,x=0.549
However Symbolab only pleases me with:
View attachment 10378

I'd be very glad if someone can explain to me what I am doing wrong
So I continue, and try solving it.

I'd be glad if someone can explain my mistake, and calm my mind. I really can't imagine symbolab bugging out on me and giving me a wrong answer.

Thanks for the help in advance
KHper3

What you are doing is a valid way to solve a quadratic: it is called "completing the square." It is summarized by the quadratic formula. I recommend that you memorize the quadratic formula and use it, not because completing the square is wrong, but rather because completing the square is slow and quite prone to error. Below, using baby steps, I use your method to derive the quadratic formula.

\(\displaystyle a \ne 0 \text { and } ax^2 + bx + c = 0 \implies\)

\(\displaystyle ax^2 + bx = -\ c \implies\)

\(\displaystyle x^2 + \left ( \dfrac{b}{a} * x \right ) = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + \left ( \dfrac{2}{2} * \dfrac{b}{a} * x \right ) = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + 2 \left ( \dfrac{b}{2a} \right ) x = -\ \dfrac{c}{a} \implies\)

\(\displaystyle x^2 + 2 \left ( \dfrac{b}{2a} \right ) x + \left ( \dfrac{b}{2a} \right )^2 = -\ \dfrac{c}{a} + \left ( \dfrac{b}{2a} \right )^2 \implies \)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a} \implies \)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \left ( \dfrac{c}{a} * \dfrac{4a}{4a} \right ) \implies\)

\(\displaystyle \left ( x + \dfrac{b}{2a} \right )^2 = \dfrac{b^2}{4a^2} - \dfrac{4ac}{4a^2} = \dfrac{b^2 - 4ac}{4a^2} \implies\)

\(\displaystyle \sqrt{\left (x + \dfrac{b}{2a} \right )^2} = \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}} = \pm \dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}} = \pm \dfrac{\sqrt{b^2- 4ac}}{2a} \implies\)

\(\displaystyle x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2-4ac}}{2a} \implies\)

\(\displaystyle x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.\)

I am not a great fan of formulas, but the quadratic formula will save you loads of time and prevent many errors.

EDIT: Now apply this to your problem.

\(\displaystyle x = \dfrac{-\ (-\ 2) \pm \sqrt{(-\ 2)^2 - 4(3)(-\ 2)}}{2 * 3} = \dfrac{2 \pm \sqrt{4 + 24}}{6} = \\

\dfrac{2 \pm \sqrt{28}}{6} = \dfrac{2 \pm \sqrt{4 * 7}}{6} = \dfrac{2 \pm 2\sqrt{7}}{6} = \dfrac{1 \pm \sqrt{7}}{3}.\)

Basically, just arithmetic. Calm your mind by taking advantage of a simple formula.

 

[KH%3]

thank you so much for taking your time- i appreciate it!
i will dig a bit into it, so i can memorize it

 

[KH%3]

i have another mistery sometimes it seems that i am able to do this:

And then other times, the almost same approach, will not give me the correct result:


It seems that I don't understand the rules yet...?

best regards,
KHper3