Solve quadratic equations using square root property

[bobisaka]

I got the answer correct once, but when I did it again I got it wrong. I have done it a myriad of different ways since, but I am doing something wrong.

the photo is my last attempt..

 

[lev888]

I got the answer correct once, but when I did it again I got it wrong. I have done it a myriad of different ways since, but I am doing something wrong.

the photo is my last attempt..

Don't convert 2 into 4/2. Just divide it by the 2 in the denominator. This is not a mistake, just an unnecessary complication.
You are not squaring the parentheses correctly. Do you know the formula for the square of a sum?

 

[Dr.Peterson]

I got the answer correct once, but when I did it again I got it wrong. I have done it a myriad of different ways since, but I am doing something wrong.

the photo is my last attempt..

Where you wrote the word "solve", you appear to be actually checking the solution you had in the line above. That's a good thing to do, but you need to know what "solve" means.

I often find that students make more mistakes in checking their work than in solving! That's often because the check is harder, in some sense. It's also because you often move on to do new stuff without having fully mastered the earlier material -- or else you've forgotten the basics.

Here, your solution is fine (though you may be expected to simplify the radical); but in the check, you forgot that the best way to evaluate an expression is usually to do exactly what it says, not to do fancy things like distributing.

Just evaluate what's inside the parentheses, and then square the result. To multiply a fraction by 2/1, you don't need a common denominator; just either multiply the top by 2 and the bottom by 1, or "cancel" the 2's. Once you've done that, it will be easy to subtract the 3; and what's left will be easy to square.

By trying to square the whole expression, you made it harder, and also made a major mistake, because the square of $a+b$ is not $a^2+b^2$.

 

[bobisaka]

Am I doing it right?

did I miss the square of a sum? I did not require that formula in this one..

 

[Dr.Peterson]

Am I doing it right?

did I miss the square of a sum? I did not require that formula in this one..

Now you've got it right. (The simplification is not need in the check, just in how teachers often expect you to write your answer.)

The square of a sum, done incorrectly, was here:

to

It's a subtraction, not a product and $(a-b)^2$ is equal to $a^2-2ab+b^2$, not to $a^2-b^2$ as you assumed.

But as I said, its quicker and safer to take the new way, in large part because you avoided having to do this error-prone task.

 

[JeffM]

I got the answer correct once, but when I did it again I got it wrong. I have done it a myriad of different ways since, but I am doing something wrong.

the photo is my last attempt..

You made numerous errors from the start.

$$(2x - 3)^2 = - 12 \text { DOES NOT RESULT IN } 2x - 3 = \sqrt{-12}.$$
The correct result is $2x - 3 = \pm \sqrt{-12}.$

It is false that $\pm \sqrt{-12} = \sqrt{12}$.

It is also false that $\sqrt{12} = \pm \sqrt{12}.$

Practically every line is in error.

But I wonder how you started with $(2x - 3)^2 = - 12.$

Are you studying complex numbers. If not, what was the original problem?

 

[bobisaka]

You made numerous errors from the start.

$$(2x - 3)^2 = - 12 \text { DOES NOT RESULT IN } 2x - 3 = \sqrt{-12}.$$
The correct result is $2x - 3 = \pm \sqrt{-12}.$

It is false that $\pm \sqrt{-12} = \sqrt{12}$.

It is also false that $\sqrt{12} = \pm \sqrt{12}.$

Practically every line is in error.

But I wonder how you started with $(2x - 3)^2 = - 12.$

Are you studying complex numbers. If not, what was the original problem?

Thanks for the clarification.

That was the example question..I neglected the plus-minus sign on that second line. I kind of used it with out fully comprehending it Algebra-ically.

Only recently learned the complex numbers.

 

[bobisaka]

Now you've got it right. (The simplification is not need in the check, just in how teachers often expect you to write your answer.)

The square of a sum, done incorrectly, was here:

View attachment 28588 to View attachment 28589

It's a subtraction, not a product and $(a-b)^2$ is equal to $a^2-2ab+b^2$, not to $a^2-b^2$ as you assumed.

But as I said, its quicker and safer to take the new way, in large part because you avoided having to do this error-prone task.

Yes, I agree. Thanks.