ANSWER:
If x = [tex]\frac{2}{3}[/tex] and x = -3 are the roots of the equation, then values of a and b are 3, -6
SOLUTION:
Given, quadratic equation is [tex]a x^{2}+7 x+b=0[/tex] and its roots are -3 , [tex]\frac{2}{3}[/tex]
We know that, for any quadratic equation of form [tex]a x^{2}+b x+c=0[/tex] with roots [tex]x_{1} and x_{2}[/tex] then,
Sum of roots [tex](x_{1} +x_{2})[/tex] = [tex]\frac{-b}{a}[/tex]
Product of roots ( [tex]x_{1} \times x_{2})[/tex] = [tex]\frac{c}{a}[/tex]
Now, for given quadratic equation [tex]x_{1}[/tex] = -3 and [tex]x_{2}[/tex] = [tex]\frac{2}{3}[/tex]
hence Sum of roots = [tex]\frac{-7}{a}[/tex]
[tex]x_{1} + x_{2} = \frac{-7}{a}[/tex]
[tex]-3 + \frac{2}{3} = \frac{-7}{a}[/tex]
on solving we get "a" = 3
Now product of roots = [tex]x_{1} \times x_{2} = \frac{b}{3}[/tex]
[tex]-3 \times \frac{2}{3} = \frac{b}{3}[/tex]
b = -6
hence, the values of a and b are 3, -6.
