Junior Algebra Math question on...

The question is:
Determine the number and type of solutions for:

X^2-4x+10=0

By examining the discriminant.

I looked up the formula for finding the discriminant of an equation:

D = b^2-4ac

So

D = (-4)^2-4(1)10
= 16-40
= -24

Correct me if I'm wrong.
The discriminant is -24, so if anyone could help me out and tell me the number and type of solutions for the equation, by examining the discriminant?

Easy 10 points if you understand the topic, unlike myself lol.

Thank you.

Seeing as you're in junior algebra, if the discriminant is negative then there is no solution to your quadratic. This is not true later on. And your working looks fine so there is no solution to your problem.

If:
D>0 then two solutions
D=0 one solution
D<0 no solutions

The equation is quadratic - you can tell this by the highest power of x =2, so it will have at most, 2 solutions.

The discriminant is negative, therefore the solution(s) will be complex.

the whole quadratic equation for finding the solution(s) is

x =( -b +/- SQRT(b^2-4*a*c) )/2

The discriminant is the stuff in the radical - taking the square root of a negative number always generates imaginary numbers.

First : Find Discriminant (D)= (16) - 4*1*10 = 16 - 40 = -24.

1- If D is positive value (D>0); Then the equation will have 2 Roots.
2- If D is ZERO (D=0); Then the equation will have 2 Identical Roots. (Which means that it only has 1 unique solution.
3-If D is negative value (D<0); Then the equation will haveNo Real Roots.

Your case follows point 3 in which D<0, So no solution exist within the Real Number Group.

Good luck...

x^2-4x+10=0
here a=1 b= -4 c=10
roots are
[-b+/-sqrt(b^2-4ac) ]/(2a)
b^2-4ac=16-40= - 24
neagtive. There is no real root for a negative number and such roots become imaginary roots.
As eqn is parabolic with x^2 in it, there will be 2 roots but both of them are imaginary.

D>0 - two real numbers
D=0 - one real number
D<0 -Complex conjugate solution ( two complex numbers : x+yj and x-yj where j is the square root of 1)

D>0 then two solutions
D=0 one solution
D<0 no solutions

It means the equation doesn's has the real answer

i got the same answer :)