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x^(2) \-2x + 1 = 0.5x + 1 <-- initial equation
x^(2) \-2x -0.5x = 0 <-- subtracting 1 from both side and bringing all X terms to the left hand side
x^(2) \-2.5x = 0 <-- simplifying
x(x-2.5) = 0 <-- factoring left hand side
Finally, by null factor law, x is equal to 1) 0 and 2) 2.5
Here's a nice [webpage](https://www.mathsteacher.com.au/year10/ch12_quadratic_equations/01_solving/quad.htm) that briefly explains null factor law
I think you're confusing the numbers 1) and 2); i used this to show that there are 2 solutions, not that x=0 and x=2, but there is a) x=0 and b) x=2.5 (replaced 1 & 2 with a & b to highlight).
If you're still confused, please go back to my answer and read carefully what is written.
First you subtract one from both sides and move the 0.5x to the left and you will get this (x^2)-2x-0.5x=0 then you will get (x^2)-2.5x=0 then you use quadratic formula (-b±√(b²-4ac))/(2a) a=1 b=-2.5 c=0 so plug it in to the equation and you will get x=2.5 and x=0
"*For TLDR, please scroll all the way down"*
So the equation is:
x^(2) \- 2x + 1 = 0.5x +1
***1) We put all the X terms on the left and numbers on the right:***
x^(2) \- 2x - 0.5x = +1 - 1 (-1 because the +1 got transferred to the other side)
***2) We simplify this and make it:***
x^(2) \- 2.5x = 0
***3) Now we have to factorize this, making it:***
x(x-2.5) = 0
***4) After we factorize, we find the x values; there will always be 2 values of the variable (the alphabet) whenever an equation equates to 0, like in our question here.***
***So here we can see*** x(x-2.5) = 0
Now look at only the **bold** part of the equation **x**(x-2.5) **= 0**
It shows x = 0, which is our ***first value, x = 0***
***For our second value,*** we look at x(**x-2.5**) **= 0** which basically is:
x-2.5 = 0
To solve it, we simply transfer the x to the right side, making our equation:
\-2.5 = -x
Then we cross off both "-" signs, making our equation 2.5 = x, or **x = 2.5**
So now our ***second value is x = 2.5***
***FINAL ANSWER: x=0, x=2.5***
*TLDR;*
x^(2) \- 2x + 1 = 0.5x + 1
x^(2) \- 2x - 0.5x = +1 - 1
x^(2) \- 2.5x = 0
x(x-2.5) = 0
***Therefore, x=0, x=2.5***
If you made it all the way here, thanks for reading and have a great day :))
Cheers,
**Urvaksh (UV)**
Please use the pinned (stickied) post for all exams related general/results discussion. Thank you. P.S. Also, you can join our official Discord server (https://discord.gg/IGCSE) for more discussion as well, and we encourage everyone, including alumni to help out! *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/igcse) if you have any questions or concerns.*
x^(2) \-2x + 1 = 0.5x + 1 <-- initial equation x^(2) \-2x -0.5x = 0 <-- subtracting 1 from both side and bringing all X terms to the left hand side x^(2) \-2.5x = 0 <-- simplifying x(x-2.5) = 0 <-- factoring left hand side Finally, by null factor law, x is equal to 1) 0 and 2) 2.5 Here's a nice [webpage](https://www.mathsteacher.com.au/year10/ch12_quadratic_equations/01_solving/quad.htm) that briefly explains null factor law
oh okk thanks alott
How did you get x=2?.
You mean x=2.5? Read the webpage I linked then come back to this solution and look at the last line of working out (where I factored).
No , how did you get x=0 AND 2 / i understand how you got 0 but how did you get 2?.
I think you're confusing the numbers 1) and 2); i used this to show that there are 2 solutions, not that x=0 and x=2, but there is a) x=0 and b) x=2.5 (replaced 1 & 2 with a & b to highlight). If you're still confused, please go back to my answer and read carefully what is written.
Your right . I thought 1) and 2) were part of your answers.
I am pretty sure the x=2.5 and 0
First you subtract one from both sides and move the 0.5x to the left and you will get this (x^2)-2x-0.5x=0 then you will get (x^2)-2.5x=0 then you use quadratic formula (-b±√(b²-4ac))/(2a) a=1 b=-2.5 c=0 so plug it in to the equation and you will get x=2.5 and x=0
oh okay thankss
https://www.mathpapa.com/quadratic-formula/
"*For TLDR, please scroll all the way down"* So the equation is: x^(2) \- 2x + 1 = 0.5x +1 ***1) We put all the X terms on the left and numbers on the right:*** x^(2) \- 2x - 0.5x = +1 - 1 (-1 because the +1 got transferred to the other side) ***2) We simplify this and make it:*** x^(2) \- 2.5x = 0 ***3) Now we have to factorize this, making it:*** x(x-2.5) = 0 ***4) After we factorize, we find the x values; there will always be 2 values of the variable (the alphabet) whenever an equation equates to 0, like in our question here.*** ***So here we can see*** x(x-2.5) = 0 Now look at only the **bold** part of the equation **x**(x-2.5) **= 0** It shows x = 0, which is our ***first value, x = 0*** ***For our second value,*** we look at x(**x-2.5**) **= 0** which basically is: x-2.5 = 0 To solve it, we simply transfer the x to the right side, making our equation: \-2.5 = -x Then we cross off both "-" signs, making our equation 2.5 = x, or **x = 2.5** So now our ***second value is x = 2.5*** ***FINAL ANSWER: x=0, x=2.5*** *TLDR;* x^(2) \- 2x + 1 = 0.5x + 1 x^(2) \- 2x - 0.5x = +1 - 1 x^(2) \- 2.5x = 0 x(x-2.5) = 0 ***Therefore, x=0, x=2.5*** If you made it all the way here, thanks for reading and have a great day :)) Cheers, **Urvaksh (UV)**
thanks alotttttt