Respuesta :
ANSWER
[tex]x = \frac{2}{3} \: or \: x = -4[/tex]
EXPLANATION
The given quadratic equation is
[tex]3 {x}^{2} + 10x + c = 0[/tex]
a=3,b=10
Let m and n be the roots of this equation.
Then, the sum of roots
[tex]m + n = \frac{ - b}{a} [/tex]
[tex]m + n = - \frac{10}{3} [/tex]
The product of roots,
[tex]mn = \frac{c}{a} [/tex]
[tex]mn = \frac{c}{3} [/tex]
The difference between the two roots
[tex]m - n = \sqrt{ {(m + n)}^{2} - 4mn } [/tex]
[tex]4 \frac{2}{3} = \sqrt{ {( - \frac{10}{3} )}^{2} - \frac{4c}{3} } [/tex]
[tex]\frac{14}{3} = \sqrt{ \frac{100}{9} - \frac{4c}{3} } [/tex]
Square both sides
[tex]\frac{196}{9} = \frac{100}{9} - \frac{4c}{3} [/tex]
Solve for c.
[tex] \frac{4c}{3}= \frac{100}{9} - \frac{196}{9 } [/tex]
[tex] \frac{4c}{3}= - \frac{96}{9}[/tex]
[tex]c= - \frac{96}{9} \times \frac{3}{4} [/tex]
[tex]c = - 8[/tex]
The quadratic equation now becomes;
[tex]3 {x}^{2} + 10x - 8 = 0[/tex]
The solution is given by,
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
[tex]x = \frac{ 10 \pm \sqrt{ {(10)}^{2} - 4(3)( - 8)} }{2(3)} [/tex]
[tex]x = \frac{ - 10 \pm \sqrt{ 100 + 96} }{6} [/tex]
[tex]x = \frac{ - 10 \pm \sqrt{ 196} }{6} [/tex]
[tex]x = \frac{ - 10 \pm 14 }{6} [/tex]
[tex]x = \frac{2}{3} \: or \: x = -4[/tex]
[tex]x = \frac{2}{3} \: or \: x = -4[/tex]
EXPLANATION
The given quadratic equation is
[tex]3 {x}^{2} + 10x + c = 0[/tex]
a=3,b=10
Let m and n be the roots of this equation.
Then, the sum of roots
[tex]m + n = \frac{ - b}{a} [/tex]
[tex]m + n = - \frac{10}{3} [/tex]
The product of roots,
[tex]mn = \frac{c}{a} [/tex]
[tex]mn = \frac{c}{3} [/tex]
The difference between the two roots
[tex]m - n = \sqrt{ {(m + n)}^{2} - 4mn } [/tex]
[tex]4 \frac{2}{3} = \sqrt{ {( - \frac{10}{3} )}^{2} - \frac{4c}{3} } [/tex]
[tex]\frac{14}{3} = \sqrt{ \frac{100}{9} - \frac{4c}{3} } [/tex]
Square both sides
[tex]\frac{196}{9} = \frac{100}{9} - \frac{4c}{3} [/tex]
Solve for c.
[tex] \frac{4c}{3}= \frac{100}{9} - \frac{196}{9 } [/tex]
[tex] \frac{4c}{3}= - \frac{96}{9}[/tex]
[tex]c= - \frac{96}{9} \times \frac{3}{4} [/tex]
[tex]c = - 8[/tex]
The quadratic equation now becomes;
[tex]3 {x}^{2} + 10x - 8 = 0[/tex]
The solution is given by,
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
[tex]x = \frac{ 10 \pm \sqrt{ {(10)}^{2} - 4(3)( - 8)} }{2(3)} [/tex]
[tex]x = \frac{ - 10 \pm \sqrt{ 100 + 96} }{6} [/tex]
[tex]x = \frac{ - 10 \pm \sqrt{ 196} }{6} [/tex]
[tex]x = \frac{ - 10 \pm 14 }{6} [/tex]
[tex]x = \frac{2}{3} \: or \: x = -4[/tex]
We will see that the two solutions are x = -4 and x = 2/3.
Solving the quadratic equation:
Here we have the quadratic equation:
3*x^2 + 10x + c = 0
The solutions are given by the Bhaskara's equation, we will get:
[tex]x = \frac{-10 \pm \sqrt{10^2 - 4*3*c} }{2*3}[/tex]
Now, we know that the difference between the roots is 4 + 2/3
Then we have:
[tex]x_+ = \frac{-10 + \sqrt{10^2 - 4*3*c} }{2*3}\\\\x_- = \frac{-10 - \sqrt{10^2 - 4*3*c} }{2*3}\\\\The\ difference\ is:\\\\D = x_+ - x_- = \frac{-10 + \sqrt{10^2 - 4*3*c} }{2*3} - \frac{-10 - \sqrt{10^2 - 4*3*c} }{2*3} = \frac{ \sqrt{10^2 - 4*3*c} }{3} = 4 + 2/3[/tex]
Now we can solve this last equation to find the value of C.
[tex]\frac{\sqrt{10^2 - 4*3*c} }{3} = 4 + 2/3 = \frac{14}{3} \\\\\sqrt{10^2 - 4*3*c} = 14\\\\10^2 - 4*3*c = 14^2 = 196\\\\-4*3*c = 196 - 100 = 96\\\\c = 96/(-4*3) = -8[/tex]
Then the equation is:
3*x^2 + 10*x - 8 = 0
And the two solutions are:
[tex]x = \frac{-10 \pm \sqrt{10^2 - 4*3*-8} }{2*3}\\\\x = \frac{-10 \pm 14 }{6}\\\\x_+ = \frac{-10 + 14 }{6} = 2/3\\\\x_- = \frac{-10 - 14 }{6}\ = - 4[/tex]
If you want to learn more about quadratic equations, you can read:
https://brainly.com/question/1214333