Answer:
Discriminant: -4
No real solutions
Step-by-step explanation:
You might remember this long, seemingly very hairy thing called the quadratic formula from earlier in algebra:
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
The discriminant of a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] is that bit under the square root: [tex]b^2-4ac[/tex]. What does the discriminant tell us? Since we're taking its square root, we know that [tex]\sqrt{b^2-4ac}[/tex] is only real for non-negative values of [tex]b^2-4ac[/tex]. If [tex]b^2-4ac > 0[/tex], we have two real roots, one for [tex]\sqrt{b^2-4ac}[/tex] and one for [tex]-\sqrt{b^2-4ac}[/tex]. If [tex]b^2-4ac = 0[/tex], the function has a rational root, since the formula becomes
[tex]x=\frac{-b}{2a}[/tex]
In your case, we have the equation [tex]x^2-4x+5[/tex]; here, [tex]a=1[/tex], [tex]b=-4[/tex], and [tex]c=5[/tex], so our discriminant is [tex](-4)^2-4(1)(5)=16-20=-4[/tex]. Since we'd have a negative under our square root in the quadratic formula, we have no real solutions.
