Question

hat is the value of the discriminant, b2 − 4ac, for the quadratic equation 0 = x2 − 4x + 5, and what does it mean about the number of real solutions the equation has?

Answer 1

`\bf \qquad \qquad \qquad \textit{discriminant of a quadratic} \\\\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{-4}x\stackrel{\stackrel{c}{\downarrow }}{+5} ~~~~~~~~ \stackrel{discriminant}{b^2-4ac}= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases} \\\\\\ (-4)^2-4(1)(5)\implies 16-20\implies -4\impliedby \textit{no solution}`

Answer 2

The discrimination value of b^2-4ac corresponds to the general equation of Ax+Bx+C=0 So for your equation x^2-4x+5=0 b=-4 so b^2=(-4)^2=16 a=1c=5 So we have (-4)^2-4(1)(5)= 16-20=-4 is the discriminant value. The number of real solutions are determined by the discriminant.b^2-4ac>0 = 2 real solutions b^2-4ac=0 = 1 real solutionb^2-4ac<0 = no real solutions. This is the category your equation false into so there are no REAL solutions. You will learn to solve these in calculus.
The number of real solutions are the number of values that satisfy the equation for x. Otherwise the values that x can be to hold the equation to be true.

The answer is B in case you don't want to read the lengthy explanation. Hope this helps. :)