Solving Quadratic Equation with Solutions x1 and x2: How to Calculate x1-3+x2-3?

In summary: I did as you said and I got:((x1+x2)/x1x2)3-3x1/x12x22 -3x2/x22x12. then I added the minuses together:((x1+x2)/x1x2)3-3x1+3x2/(x1x2)2Did I do this part correct?May I also ask, what command did you use to write (x1+x2)/x1x2 because your way looks much easier to understandthank you for your reply. I did as you said and I got:(x1+x2/x
  • #1
Matejxx1
72
1
1.given the equation a(3x2+2x+1)=4x-6x2-4 with the solutions x1 and x2
a) let a=0 without solving the equation calculate x1-3+x2-3

the correct answer is supposed to be -7/2


Homework Equations


x1+x2=-b/a
x1*x2=c/a




The Attempt at a Solution


The first thing I was that I put in "a" so,
0*(3x2+2x+1)=4x-6x2-4
0=4x-6x2-4...... here I multiplied the whole equation by -1
0=6x2-4x+4......here i divided the equation by 6
0=x2-4/6x+46
then I used the equation
1/x13+1/x23=(1/x1+1/x2)3 -3/x12x2 -3/x22x1
This is where I have no idea how to continue.I'm not even sure that what I have written is correct .If someone could help me solve this,I would really appreciate it. Thanks for reading .
 
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  • #2
Matejxx1 said:
1.given the equation a(3x2+2x+1)=4x-6x2-4 with the solutions x1 and x2
a) let a=0 without solving the equation calculate x1-3+x2-3

the correct answer is supposed to be -7/2


Homework Equations


x1+x2=-b/a
x1*x2=c/a




The Attempt at a Solution


The first thing I was that I put in "a" so,
0*(3x2+2x+1)=4x-6x2-4
0=4x-6x2-4...... here I multiplied the whole equation by -1
0=6x2-4x+4......here i divided the equation by 6
0=x2-4/6x+4/6
then I used the equation
1/x13+1/x23=(1/x1+1/x2)3 -3/x12x2 -3/x22x1
This is where I have no idea how to continue.I'm not even sure that what I have written is correct .If someone could help me solve this,I would really appreciate it. Thanks for reading .

Try to bring in x1+x2 and x1x2.
For example, [tex]1/x_1+1/x_2=\frac{x_1+x_2}{x_1 x_2}[/tex]

ehild
 
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  • #3
thank you for your reply. I did as you said and I got:
((x1+x2)/x1x2)3-3x1/x12x22 -3x2/x22x12. then I added the minuses together:
((x1+x2)/x1x2)3-3x1+3x2/(x1x2)2
Did I do this part correct?
May I also ask, what command did you use to write (x1+x2)/x1x2 because your way looks much easier to understand
 
Last edited:
  • #4
Matejxx1 said:
thank you for your reply. I did as you said and I got:
(x1+x2/x1x2)3-3x1/x12x22 -3x2/x22x12. then I added the minuses together:
(x1+x2/x1x2)3-3x1+3x2/(x1x2)2
Did I do this part correct?
May I also ask, what command did you use to write x1+x2/x1x2 because your way looks much easier to understand

He did not write ##x_1+x_2/x_1x_2##; you did that. In fact, you wrote
[tex]x_1 + \frac{x_2}{x_1 x_2}[/tex]
If you mean
[tex] \frac{x_1 + x_2}{x_1 x_2}[/tex]
then either use LaTeX or else use parentheses, like this: (x1+x2)/x1x2.
 
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Likes 1 person
  • #5
Ray Vickson said:
He did not write ##x_1+x_2/x_1x_2##; you did that. In fact, you wrote
[tex]x_1 + \frac{x_2}{x_1 x_2}[/tex]
If you mean
[tex] \frac{x_1 + x_2}{x_1 x_2}[/tex]
then either use LaTeX or else use parentheses, like this: (x1+x2)/x1x2.
Thank you I didn't notice that and I fixed it right away
 
  • #6
Now I added in the numbers
x1+x2=4/6
x1*x2=4/6
(((4/6)/4/6))-((3*4/6)/(4/6)2 =
1-(2)/0,444=
1-4,5=-3,5=-7/2
Thank you very much for your help I got the right answer
 

1. What is the general formula for solving a quadratic equation?

The general formula for solving a quadratic equation is x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.

2. How do I calculate the discriminant of a quadratic equation?

The discriminant of a quadratic equation is the value inside the square root in the general formula. It is calculated as b^2 - 4ac.

3. What is the significance of x1 and x2 in the solutions of a quadratic equation?

The solutions x1 and x2 represent the two possible values of x that satisfy the quadratic equation. These values are also known as the roots of the equation.

4. Can I solve a quadratic equation with complex solutions?

Yes, a quadratic equation can have complex solutions if the discriminant is negative. In this case, the solutions will be in the form a + bi, where a and b are real numbers and i is the imaginary unit.

5. Is there a specific method for calculating x1-3+x2-3 in a quadratic equation?

No, there is no specific method for calculating x1-3+x2-3. This expression is simply the sum of the two solutions of the quadratic equation. It is important to note that x1 and x2 should be calculated separately before adding them together.

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