Question

If $x = \dfrac{2}{3}$ and x = -3 are the roots of the equation $a{x^2} + 7x + b = 0,$ find the values of ${a^2} + {b^2}.$

Answer 1

Hint: Factorization of a quadratic equation gives us roots. Here the roots are given. We can follow the reverse process to find the required quadratic equation and then compare it with the given equation to get coefficients a, b.

Complete step-by-step answer:
The given quadratic equation is $a{x^2} + 7x + b = 0$

The roots of this equation are given as $x = \dfrac{2}{3}\;\& \;x = - 3$.

Then we can write as
$\left( {x - \dfrac{2}{3}} \right)\left( {x - ( - 3)} \right) = 0$ It will be the quadratic equation found from the given roots.
$ \Rightarrow \left( {x - \dfrac{2}{3}} \right)\left( {x + 3} \right) = 0$

On simplification of the above equation,
$ \Rightarrow {x^2} + 3x - \dfrac{{2x}}{3} - \dfrac{{2 \times 3}}{3} = 0$
$ \Rightarrow {x^2} + \dfrac{7}{3}x - 2 = 0$
Multiplying the above equation with 3 on both sides, we get
$ \Rightarrow 3{x^2} + 7x - 6 = 0$
Comparing the above equation with the given quadratic equation $a{x^2} + 7x + b = 0,$ we get
a = 3, b = –6.
$ \Rightarrow {a^2} + {b^2} = {(3)^2} + {( - 6)^2} = 9 + 36 = 45$
$\therefore $ The value of ${a^2} + {b^2} = 45$

Note: We can use a different way to solve the given problem using sum of the roots and product of the roots. Standard form of a quadratic equation with roots a, b can be written as
${x^2} - (sum\;of\;roots)x + product\;of\;roots = 0$
${x^2} - (a + b)x + ab = 0$