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Mathematics 63 Online
OpenStudy (anonymous):

Write the value of the discriminant of each equation. Then use it to decide how many different real-number root the equation has. (do not solve.) x^2-3x+2

OpenStudy (anonymous):

the discriminant is the "under the square root" stuff in the quadratic formula. \[\sqrt{b^2-4ac}\]

OpenStudy (anonymous):

What I don't get is, when I put it into the discriminant: it becomes 9-8... which is 1.. and I have no idea where that goes or what proves what.

OpenStudy (anonymous):

ok. yeah

OpenStudy (anonymous):

HAHA. :)

OpenStudy (anonymous):

it is =1. this tells you how many roots you have. \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]If the discriminant is negative, you have no solutions/root (cant square root a negative number). If the discriminant is =0, the you have one real solution. If it is positive, you have two real roots.

OpenStudy (anonymous):

OHHHHHH I SEEEE HAAHHAHAHAHH what i DIDN'T know was you had to plug it into the quaratic formula. now I see! so ... I have 2 real solutions right?

OpenStudy (campbell_st):

the conditions for the discriminant are \[b^2 - 4ac > 0\] two unequal real roots...if its a perfect square the roots are rational if not irrational \[b^2 - 4ac = 0\] 2 equal real roots.... this is the perfect square case (x + b)^2 = 0 \[b^2 - 4ac < 0\] two inequal complex roots.... or they made be described as no real roots.

OpenStudy (anonymous):

so.... i'm right right?..

OpenStudy (anonymous):

yep; 2 roots.

OpenStudy (anonymous):

technically cambell st is correct with the =0 case. I always say one real root but it is actually 2 equal real roots when discriminant =0.

OpenStudy (anonymous):

haah thanks :D

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