Question

The difference between the two roots of the equation 3x2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

Answer 1

ANSWER

$x = \frac{2}{3} \: or \: x = -4$

EXPLANATION

The given quadratic equation is

$3 {x}^{2} + 10x + c = 0$

a=3,b=10

Let m and n be the roots of this equation.

Then, the sum of roots

$m + n = \frac{ - b}{a}$

$m + n = - \frac{10}{3}$

The product of roots,

$mn = \frac{c}{a}$

$mn = \frac{c}{3}$

The difference between the two roots

$m - n = \sqrt{ {(m + n)}^{2} - 4mn }$

$4 \frac{2}{3} = \sqrt{ {( - \frac{10}{3} )}^{2} - \frac{4c}{3} }$

$\frac{14}{3} = \sqrt{ \frac{100}{9} - \frac{4c}{3} }$

Square both sides

$\frac{196}{9} = \frac{100}{9} - \frac{4c}{3}$

Solve for c.

$\frac{4c}{3}= \frac{100}{9} - \frac{196}{9 }$

$\frac{4c}{3}= - \frac{96}{9}$

$c= - \frac{96}{9} \times \frac{3}{4}$

$c = - 8$

The quadratic equation now becomes;

$3 {x}^{2} + 10x - 8 = 0$

The solution is given by,

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

$x = \frac{ 10 \pm \sqrt{ {(10)}^{2} - 4(3)( - 8)} }{2(3)}$

$x = \frac{ - 10 \pm \sqrt{ 100 + 96} }{6}$

$x = \frac{ - 10 \pm \sqrt{ 196} }{6}$

$x = \frac{ - 10 \pm 14 }{6}$

$x = \frac{2}{3} \: or \: x = -4$