Question

If $x = 1$ is a common root of the quadratic equation $a{x^2} + ax + 3 = 0$ and ${x^2} + x + b = 0$ , then determine the value of $ab$ .

Answer 1

Here we are asked to find the value of the product $ab$ from the given data. For this we need the values of $a\& b$ so, we are given one of the roots of the given equation so we will assume the other root to be any variable and substitute it in the equation as it satisfies it. Then using that substituted equation we will solve to find the value of $a\& b$ then we will find the required product

Complete answer:
We have two quadratic equations given by $a{x^2} + ax + 3 = 0$ and ${x^2} + x + b = 0$ .
It is also given that $x = 1$ is a common root of these quadratic equations.
Since, any root of the quadratic equation satisfies the given equation so $x = 1$ can be substituted in the given quadratic equations.
Substitute $x = 1$ in the quadratic equation $a{x^2} + ax + 3 = 0$ ,
$a{(1)^2} + a(1) + 3 = 0$
$a + a + 3 = 0$
Solve the above equation for $a$ ,
$2a + 3 = 0$
Subtract 3 from both sides of the above equation
$2a + 3 - 3 = 0 - 3$
Simplify,
$2a = - 3$
Divide $2$ from both sides of the equation
$\dfrac{{2a}}{2} = - \dfrac{3}{2}$
Simplify the above equation,
$a = - \dfrac{3}{2}$
Substitute $x = 1$ in the quadratic equation ${x^2} + x + b = 0$ ,
${(1)^2} + (1) + b = 0$
$1 + 1 + b = 0$
Solve the above equation for $b$ ,
$2 + b = 0$
Subtract $2$ from both sides of the equation,
$2 + b - 2 = 0 - 2$
Simplify the above equation,
$b = - 2$
Now, we have found the values of $a$ and $b$ ,
$a = - \dfrac{3}{2}$ and $b = - 2$
Determine the value of $ab$ by substituting the above values of $a$ and $b$ in $ab$ ,
$ab = - \dfrac{3}{2} \times ( - 2)$
Cancelling the $2$ ,
$ab = - 3 \times ( - 1)$
Since the product of two negative numbers is a positive number, so we get the solution as $3$ that is
$ab = 3$
If $x = 1$ is a common root of the quadratic equation $a{x^2} + ax + 3 = 0$ and ${x^2} + x + b = 0$ , then determine the value of $ab$ is $3$ .

Note:
A quadratic equation which is given by $a{x^2} + bx + c = 0$ , such that the sum of the values of $a$ , $b$ and $c$ is $0$ , that is $a + b + c = 0$ , then one of the root of the given quadratic is definitely $1$ . It can also be checked in the reverse order. If the root of the quadratic equation is $1$ so it satisfies the equation, then by substituting $x = 1$in the given quadratic equation, we get $a + b + c = 0$ .