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
the discriminant is the "under the square root" stuff in the quadratic formula. \[\sqrt{b^2-4ac}\]
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.
ok. yeah
HAHA. :)
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.
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?
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.
so.... i'm right right?..
yep; 2 roots.
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.
haah thanks :D
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