## some.random.cool.kid one year ago Solve x2 − 5x + 1 = 0 using the quadratic formula.

1. some.random.cool.kid

step one is to -1 by both sides?

2. Nnesha

now just use the quadraitc formula

3. Nnesha

no**

4. Nnesha

you already know the formula so just plugin a ,b ,c values

5. some.random.cool.kid

yeah but first -1 to the 1 and 0 go first dont they i have to simplify them.

6. Michele_Laino

comparing your equation with the general quadratic equation: $a{x^2} + bx + c = 0$ we get: a=1, b=-5, and c=1 now, please apply this formula: $x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}$

7. Nnesha

no quadratic equation should be in standard form $\huge\rm Ax^2+Bx+C=0$ equal to zero

8. Nnesha

that was for completing the square method but when we use quadratic formula we ned standard form

9. some.random.cool.kid

ok

10. some.random.cool.kid

so what is the standard is it like ... idk actually

11. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha no quadratic equation should be in standard form $\huge\rm Ax^2+Bx+C=0$ equal to zero $$\color{blue}{\text{End of Quote}}$$ ^^that's the standard form

12. Nnesha

where a=leading coefficient b=middle term c=constant term

13. Nnesha

now just plugin a ,b, c values into the quadratic formula

14. some.random.cool.kid

ok

15. some.random.cool.kid

writing all these down just in case your wondering... very helpful info for big exam EOC..

16. Nnesha

that's Awesome take ur time write theses down and practice as much as u can

17. some.random.cool.kid

k so what do i next because i wasn't sure...

18. Nnesha

did you substitute a ,b c for their values ?

19. some.random.cool.kid

a is x^2 right ?

20. some.random.cool.kid

and b is 5x?

21. Nnesha

no just the coefficient not the variable

22. Nnesha

look at the Michele'z comment a=1 bec x^2 is same as 1x^2

23. some.random.cool.kid

so 5 since its not a variable?

24. some.random.cool.kid

/so your saying 1^2? i dont get how thats possible...

25. Nnesha

yes 5 is b

26. Nnesha

there is invisible one at front of x^2 x^2 is same as 1x^2 so leading coefficient is one

27. Nnesha

i'm not saying x =1 leading coefficient is one

28. Nnesha

$\huge\rm \color{red}{1}x^2+\color{blue}{5}x+\color{green}{1}=0$ $\huge\rm \color{ReD}{A}x^2+\color{blue}{B}x+\color{green}{C}=0$

29. some.random.cool.kid

alright

30. Nnesha

oky now just plugin

31. some.random.cool.kid

k thanks i gtg thanks for your help. :D

Find more explanations on OpenStudy