Compute the discriminant. Then determine the number and type of solutions for the given equation.?
x2 - 4x + 3 = 0
a. 0; one real solution
b. -28; no real solution
c. 4; one real solution
d. 4; two unequal real solutions
x2 - 4x + 3 = 0
a. 0; one real solution
b. -28; no real solution
c. 4; one real solution
d. 4; two unequal real solutions
?
Favorite Answer
I don't know why they wanted you to use the discriminant since it is far easier to just factorise it to (x - 3)(x - 1) = 0, so x = 1 or x = 3 i.e. 2 real distinct (i.e. unequal) solutions, but since that's what they asked for, Discriminant = "b^2 - 4ac" = (-4)^2 - 4*1*3 = 16 - 12 = 4. This is > 0, so there are two distinct (unreal) roots - whole numbers in fact, although the discriminant won't tell you that.
Aleko
The answer here is d. Discriminant=4 and two unequal solutions
x=(4(+/-)sqrt(16-12))/2=3 and x=1 Just for you to remember, the discriminant is whatever is UNDER the square root!
Aminu
Using b2-4ac,
b=-4. A=1, c=3.
Then b2-4ac becomes
(-4)^2-4(1)(3)
=16-12
=4
since b2-4ac is >0, the roots are not equal.
So the answer is D.