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KJ4UTS
 one year ago
Use the discriminant to determine the number of real roots of...
x^25x7=0
x^22x+1=0
Choices are:
0
1
2
KJ4UTS
 one year ago
Use the discriminant to determine the number of real roots of... x^25x7=0 x^22x+1=0 Choices are: 0 1 2

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2what is discriminant ? do you know ?

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.1Well my work is (I just want to check): x^25x7=0 5^24(1)(7)=3>0 2 roots x^22x+1=0 2^24(1)(1)=8<0 1 root

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.1x^25x7=0 2 roots x^22x+1=0 1 root

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\(\huge\color{reD}{\rm b^24ac}\) `Discriminant` you can use this to find if the equation is factorable or not if ` b^24ac > 0` then there are 2 real zeros if ` b^24ac = 0` then there is one real root if ` b^24ac < 0` then you will get two complex roots (no xintercept)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2you should use discriminant

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{blue}{\text{Originally Posted by}}\) @KJ4UTS Well my work is (I just want to check): x^25x7=0 5^24(1)(7)=3>0 2 roots x^22x+1=0 2^24(1)(1)=8<0 1 root \(\color{blue}{\text{End of Quote}}\) b^2 so it wohuld be (2)^2 4(1)(1)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2no (2)^2 is same as 2 times 2

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2or in other words when you take `even` power of negative exponent you will get positive answer always !

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2no \[\color{ReD}{(2)^2}4(1)(1)\] i'm just talkign about (2)^2

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{blue}{\text{Originally Posted by}}\) @Nnesha \(\huge\color{reD}{\rm b^24ac}\) `Discriminant` you can use this to find if the equation is factorable or not if ` b^24ac > 0` then there are 2 real zeros if ` b^24ac = 0` then there is one real root if ` b^24ac < 0` then you will get two complex roots (no xintercept) \(\color{blue}{\text{End of Quote}}\) now read this when b^24ac =0 how many roots will u get ?

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.1x^22x+1=0 2^24(1)(1)=0=0 Which would still be 1 root?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2now what about first one ? b^24ac = ??

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.1Was my work right 5^24(1)(7)=3>0 2 roots

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2no that's not correct ....

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.1I think what I was doing wrong for both was not putting parenthesis (5)^2 for both problems into my calculator but 53>0 therefore I think it is still 2 roots?

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.1Ok thank you @Nnesha for your time and help, and for also pointing out where I went wrong :)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2np and yes you should put the parentheses (2)^2 like this :=) good job!
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