Answer:
Option C is correct.
roots of the given equation , [tex]x =\frac{2\pm i\sqrt{26}}{3}[/tex]
Step-by-step explanation:
Given the equation: [tex]3x^2+10 = 4x[/tex]
We can write this equation as:
[tex]3x^2-4x + 10 =0[/tex]
A quadratic equation is in the form of [tex]ax^2+bx+c =0[/tex] ......[1] where a,b ,c are the coefficient and x is the variable,
the solution of the equation is given by;
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
On comparing given equation with equation [1] we get;
a = 3 , b = -4 and c =10
So, the solution of the given equation is given by;
[tex]x = \frac{-(-4)\pm\sqrt{(-4)^2-4(3)(10)}}{2(3)}[/tex]
or
[tex]x = \frac{4\pm\sqrt{(16-120}}{6} = \frac{4\pm\sqrt{(-104)}}{6} = \frac{4\pm\sqrt{(-4 \times 26)}}{6}[/tex]
or
[tex]x =\frac{4\pm2 \sqrt{(-26)}}{6} = \frac{4\pm2 i\sqrt{26}}{6}[/tex] [∴[tex]\sqrt{-1} = i[/tex]
Simplify:
[tex]x =\frac{2\pm i\sqrt{26}}{3}[/tex]
therefore, the roots of the given equation are; [tex]x =\frac{2\pm i\sqrt{26}}{3}[/tex]
