Question

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

Answer 1

Answer:

x1= 2/3, x2 = -4

Step-by-step explanation:

3x² + 10x + c = 0

Formula for the roots of the quadratic equation is

$x = \frac{-b +/-\sqrt{b^2-4ac} }{2a}$

Where  a = 3, b=10, c=c  for our equation 3x² + 10x + c = 0.

$x=\frac{-10+/-\sqrt{10^{2}-4*3c} }{2*3} \\\\x=\frac{-10+/-\sqrt{100-12c} }{6} \\\\x_{1} =\frac{-10+\sqrt{100-12c} }{6} \\\\x_{2} =\frac{-10-\sqrt{100-12c} }{6} \\\\x_{1}-x{_2}=\frac{-10+\sqrt{100-12c} }{6} -(\frac{-10-\sqrt{100-12c} }{6} )\\\\x_{1}-x{_2}= \frac{2\sqrt{100-12c} }{6} =\frac{\sqrt{100-12c} }{3} \\\\x_{1}-x{_2}=\frac{\sqrt{100-12c} }{3} = 4\frac{2}{3} =\frac{14}{3} \\\\\sqrt{100-12c} =14\\(\sqrt{100-12c} )^{2}=14^{2}100-12c = 196\\12c=100-196\\c=-8$

$x=\frac{-10+/-\sqrt{100-12(-8)} }{6} = \frac{-10+/-\sqrt{196} }{6} =\frac{-10+/-14}{6} \\\\x_{1} =\frac{4}{6} =\frac{2}{3} \\x_{2} =\frac{-24}{6} =-4$