Answer:
0 real solutions
Step-by-step explanation:
The discriminant (D) is given by the formula:
[tex]D=b^2-4ac[/tex]
Where a,b, and c are gotten from the standard form of a quadratic equation, which is [tex]ax^2+bx+c=0[/tex]
The equation given is [tex]3x^2-5x+4=0[/tex]
From this we can say that a = 3, b = -5, & c = 4. Plugging these into the discriminant formula we get:
[tex]D=b^2-4ac\\D=(-5)^2-4(3)(4)\\D=-23[/tex]
There are 3 things that we can find from the value of discriminant:
1. If D=0, there are 2 equal, real roots
2. if D>0, there are 2 distinct real roots
3. if D<0, there are no real roots, rather 2 imaginary roots
Since the value of the discriminant is negative (D<0), we can say that there are no real solutions for the equation given.
Answer:
Step-by-step explanation:
[tex]\text{Use the discriminant:}\\\\ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{If}\ \Delta>0\ \text{then the equation has two real solutions:}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}\\\text{If}\ \Delta=0\ \text{then the equation has one real solution:}\ x=\dfrac{-b}{2a}\\\text{If}\ \Delta<0\ \text{then the equation has no solution.}[/tex]
[tex]\text{We have the equation:}\\\\3x^2-5x+4=0\\\\a=3,\ b=-5,\ c=4\\\\\Delta=(-5)^2-4(3)(4)=25-48=-23<0\\\\\text{no solution}[/tex]
