# If a function has a discriminant that is less than zero, what does this mean?

embizze | High School Teacher | (Level 2) Educator Emeritus

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If a quadratic function is of the form `y=ax^2+bx+c` then the discriminant is `D=b^2-4ac` .

Recall the quadratic formula -- if the quadratic is in standard form then the solutions are `x=(-b+-sqrt(b^2-4ac))/(2a)` . This implies that the solutions are `sqrt(b^2-4ac)` away from `-b/(2a)` . Graphically, if the solutions are real, then they are equidistant from the axis of symmetry `x=-b/(2a)` .

What happens when `sqrt(b^2-4ac)=0` ? The solution lies on the axis of symmetry -- thus the solution is the vertex. (This is a double root.)

What happens when `sqrt(b^2-4ac)<0` ? The left side is not real -- when you add or subtract an imaginary number from a real number you get an imaginary number. So the solutions are imaginary. Graphically, the parabola does not intersect the x-axis.

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If the discriminant is less than zero, then the quadratic has no real solutions -- it will have two imaginary solutions.

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fashionableb1 | Student, Grade 10 | (Level 1) Salutatorian

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It means that the problem has no real solutions

pramodpandey | College Teacher | (Level 3) Valedictorian

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Let us consider a quadratic function f(x)=ax^2+bx+c , a is not equals to zero f(x)=a{ x^2+(b/a)x+(c/a)} =a{(x+(b/(2a)))^2+(c/a)-(b/(2a))^2} {f(x)-(c/a)+(b/(2a))^2}=a(x+(b/(2a)))^2 {f(x)-(c/a)+(b/(2a))^2}/a =(x+(b/(2a)))^2 (i) for quadratic function important term is {f(x)-(c/a)+(b/(2a))^2}/a , not discriminant D=sqrt(b^2-4ac),what usually used in quadratic equations. The zeros of Quadratic function is roots of equation, if we solve by equating function to zero . i.e zero of function f(x) and roots of f(x)=0 are same (if exist). Zero of function y=f(x) means the point where it meet x-axis . If zero of y=f(x) does not exist , it has simple meaning it does not meet x-axis at all .