Answer:
[tex]\text{The roots of }3x^2+5x-8=0\text{ are }x=1,\frac{-8}{3}[/tex]
Step-by-step explanation:
[tex]\text{Part A: Given a quadratic equation }3x^2-15x+20=0[/tex]
[tex]\text{Comparing above equation with }ax^2+bx+c=0[/tex]
a=3, b=-15, c=20
Discriminant can be calculated as
[tex]D=b^2-4ac[/tex]
[tex]D=(-15)^2-4(3)(20)=225-240=-15<[/tex]
The roots are imaginary
The solution is
[tex]x=\frac{-b\pm\sqrt{D}}{2a}[/tex]
[tex]x=\frac{-(-15)\pm \sqrt{-15}}{2(3)}=\frac{15\pm\sqrt{15}i}{6}[/tex]
The roots are not real i.e these are imaginary
[tex]\text{Part B: Given a quadratic equation }3x^2+5x-8=0[/tex]
[tex]\text{Comparing above equation with }ax^2+bx+c=0[/tex]
a=3, b=5, c=-8
Discriminant can be calculated as
[tex]D=b^2-4ac[/tex]
[tex]D=(5)^2-4(3)(-8)=25+96=121>0[/tex]
The roots are real
By quadratic formula method
The solution is
[tex]x=\frac{-b\pm\sqrt{D}}{2a}[/tex]
[tex]x=\frac{-5)\pm \sqrt{121}}{2(3)}=\frac{-5\pm 11}{6}[/tex]
[tex]x=1,\frac{-8}{3}[/tex]
which are required roots.
I choose this method because I can get the solutions directly by substituting the values in formula, and I don't have to guess the possible solutions.
