+0

# help pls !

-1
85
6

Neil and Chris were trying to solve the quadratic equation

x^2 + bx + c = 0.

Neil wrote down the wrong value of b (but his value of c was correct), and found the roots to be 1 and 6. Chris wrote down the wrong value of c (but his value of b was correct), and found the roots to be -1 and -4. What are the actual roots of x^2 + bx + c = 0?

Mar 20, 2020

#1
+3

*****    edited   ****    edited *****

Neil    found   1 and 6      this means  (x-1)(x-6)         x^2 - 7x +  6

Chris  found   -1 and -4    this means  (x+1)(x+4)   =  x^2 + 5x + 6

If the correct values are in red     then    the equation must be    x^2 + 5x +   6        which factors to (x+3)(x+2)    roots  -3 and - 2

Mar 20, 2020
edited by ElectricPavlov  Mar 20, 2020
#2
+1

thanks !! but i entered 2 and 3 as a list, and it marked me wrong. do you know what's up?

matthewmacdell  Mar 20, 2020
#3
+2

Sorry...made a basic mistake in my answer,,,,corrected answer above Mar 20, 2020
edited by ElectricPavlov  Mar 20, 2020
#4
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Sorry the roots should be -3 and -2 I believe Pavlov

Mar 20, 2020
#5
0

YOU are right....I edited my answer above to correct my mistake !!!!

ElectricPavlov  Mar 20, 2020
#6
+1

The final roots should be -3, -2, 2, and 3!

CalTheGreat  Mar 21, 2020