4x^2 + 6x +1 = 0

J

Jhar_1986

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Aug 7, 2021
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Hi, I am studying quadratics and have a question regrading the solutions for the equation:

4x^2 + 6x +1 = 0 (1)

If I substitute either of the solutions for x into the above, the sum is not 0 ..... how can this be?

X = (-3/4)+SQRT(5)

Or

X = (-3/4)-SQRT(5)

Substituting either of these solutions for x into equation (1)

4x^2 + 6x +1 DOES NOT = 0

so how can it be explained that when y is = 0, these solutions are true?

Thanks in advance for any guidance.
James
 
Jhar_1986 said:
Hi, I am studying quadratics and have a question regrading the solutions for the equation:

4x^2 + 6x +1 = 0 (1)

If I substitute either of the solutions for x into the above, the sum is not 0 ..... how can this be?

X = (-3/4)+SQRT(5)

Or

X = (-3/4)-SQRT(5)

Substituting either of these solutions for x into equation (1)

4x^2 + 6x +1 DOES NOT = 0

so how can it be explained that when y is = 0, these solutions are true?

Thanks in advance for any guidance.
James
Click to expand...
Please check your calculations again - and share your work.

I do NOT get the solutions of x to be [-3/4 ± √5].

We need to see your work to catch your mistake.
 
Jhar_1986 said:
Hi, I am studying quadratics and have a question regrading the solutions for the equation:

4x^2 + 6x +1 = 0 (1)

If I substitute either of the solutions for x into the above, the sum is not 0 ..... how can this be?

X = (-3/4)+SQRT(5)

Or

X = (-3/4)-SQRT(5)

Substituting either of these solutions for x into equation (1)

4x^2 + 6x +1 DOES NOT = 0

so how can it be explained that when y is = 0, these solutions are true?

Thanks in advance for any guidance.
James
Click to expand...
What you've done here is to get a pair of solutions and check them. That's good.

The result of the check was to show that your answers were wrong.

So what you do now is to go back through your work and find the error, rather than to wonder why mathematics suddenly failed!

And what you'll find is a simple mistake that amounts to putting parentheses or a fraction bar in the wrong place at some step.

So go find it. Any mistake is a learning opportunity!
 
Hi James. There are different possibilities. Were those answers given to you? Maybe the expression is not formatted properly. If you goofed, that's okay. We learn by fixing mistakes.

Your materials ought to show in the quadratic formula that both the -b term and the radical term get divided by 2a.

?
 
Thanks everyone, I have been studying all day and have just seen my mistake on the thread. I was also referencing "b" as 3 as opposed to 6 in my calculations for some reason ..... Long day, glad I have not gone insane ? thanks for the support. I have never been so happy to see a zero on my calculator ha
 
(-3+SQRT(5))/4
Or
(-3-SQRT(5))/4

I think that's the right format ..... at least on maxima! I much prefer my pen. Thanks again all.
 
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