Question

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

Answer 1

Hint: As roots of the quadratic equation are given just calculate the sum of roots and product of roots with the help of the given values in the equation and roots. Then, calculate the values of a and b.

Complete step-by-step answer:
As, the given quadratic equation is \[a{x^2} + 7x + b = 0\] which is in the form \[a{x^2} + bx + c = 0\] as we have to calculate the value of a and b given in the equation.
The roots of the above equation are given as \[{x_1} = \dfrac{2}{3}\] and \[{x_2} = - 3\]
So, it will satisfy the sum of roots as \[\dfrac{{ - b}}{a}\] and the product of roots as \[\dfrac{b}{a}\] .
Here, we will first calculate sum of roots that is \[{x_1} + {x_2} = \dfrac{{ - b}}{a}\]
Substituting the values of \[{x_1}\] and \[{x_2}\] in above and also put b = 7.
We get, \[\dfrac{2}{3} - 3 = \dfrac{{ - 7}}{a}\]
By taking L.C.M on left hand side we get,
$\Rightarrow$ \[\dfrac{{2 - 9}}{3} = \dfrac{{ - 7}}{a}\]
On further simplifying:
$\Rightarrow$ \[\dfrac{{ - 7}}{3} = \dfrac{{ - 7}}{a}\]
Cancelling -7 from L.H.S and R.H.S
We get, \[\dfrac{1}{3} = \dfrac{1}{a}\]
By cross multiplying we get,
$\Rightarrow$ \[a = 3\]
Now, we will calculate the product of roots that is \[{x_1} \times {x_2} = \dfrac{b}{a}\]
Substituting the values of \[{x_1}\] and \[{x_2}\] in above and also put a = 3.
We get, \[\dfrac{2}{3} \times 3 = \dfrac{b}{3}\]
Cancelling 3 from both numerator and denominator in L.H.S.
Here,
$\Rightarrow$ \[2 = \dfrac{b}{3}\]
By cross multiplying we get,
$\Rightarrow$ \[b = 6\]
Therefore, values of a and b are 3 , 6 .

Note: In these types of questions two roots are given and a quadratic equation. As, these roots simply mean to calculate the sum of roots that is \[\dfrac{{ - b}}{a}\] and product of roots that is \[\dfrac{b}{a}\]. By calculating these two values you can easily calculate the required values which were asked in the question.