SOLUTION: Can someone please help me with my homework? Thanks :D
1.What if I have +√3 and -√3 as the roots. What is the equation?
2. Graph the equation in a sign graph: x^2+8x-
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Question 982479: Can someone please help me with my homework? Thanks :D
1.What if I have +√3 and -√3 as the roots. What is the equation?
2. Graph the equation in a sign graph: x^2+8x-7=0
In question no. 2, I find the roots of it first. Since it's not a perfect square trinomial, and 8 is an even number, I used completing the square.
x^2+8x=7
x^2+8x+(8/2)^2=7+(8/2)^2
x^2+8x+16=23
x^2+8x+16-23=0
x^2+8x-7=0
But i ended up returning to the original equation :/
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
x^2+8x=7
x^2+8x+(8/2)^2=7+(8/2)^2
x^2+8x+16=23
x^2+8x+16=23
(x+4)^2 = 23
(x+4) = +/-sqrt(23)
x= -4 +/- sqrt(23)
sum of roots = 0
product of roots = -3
x^2-(sum of roots)x+(product of roots =0
x^2-3=0 is the equation
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
1. What if I have and as the roots. What is the equation?
The equation is = , or (which is the same) = .
2. In question no. 2
See how completing the square works.
= ,
- = ,
- = ,
- = ,
= + ,
= ,
= +/- ,
= +/- .
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