Question

Write the quadratic equation whose roots are $ - 2$ and $ - 3$.

Answer 1

Hint: Here, in the given question we are given roots of quadratic equation and we need to find the equation. Quadratic equations are second-degree algebraic expressions and are of the form $a{x^2} + bx + c = 0$ where $a \ne 0$. As we know the formula for quadratic polynomial is given by ${x^2} - \left( {sum{\text{ }}of{\text{ }}zeros} \right)x - \left( {product{\text{ }}of{\text{ }}zeros} \right) = 0$ . As we can see, here coefficient of ${x^2}$ is unity so we will try to make coefficient of ${x^2}$ unity in the general form of quadratic equation $a{x^2} + bx + c = 0$.
After this, we will simplify the equation and apply conditions for sum of roots and products. By this we will find the value of variables and substitute these values back into the equation. By following these steps we will obtain our required equation.
Formula for sum of roots and product of roots is given as: product of zeros
\[\alpha + \beta = - \dfrac{b}{a}\]
$\alpha .\beta = \dfrac{c}{a}$

Complete step by step answer:
Given: Roots of the quadratic equation are $ - 2$ and $ - 3$.
Let the two roots be denoted by variables $\alpha $, $\beta $.
Let the quadratic equation be: $a{x^2} + bx + c = 0....\left( i \right)$.
Now, we will try to make the coefficient of ${x^2}$ unity by dividing the equation by $a$ on both sides.
$\dfrac{{a{x^2} + bx + c}}{a} = \dfrac{0}{a}$
It can also be written as:
$\dfrac{{a{x^2}}}{a} + \dfrac{{bx}}{a} + \dfrac{c}{a} = \dfrac{0}{a}$
By simplifying the above equation, we can write it as:
$ \Rightarrow {x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0......\left( {ii} \right)$
By basic knowledge of equation, we know that for a quadratic equation, we have relations such as:
Sum of the roots of equation $\left( i \right)$, is equal to $ - \dfrac{b}{a}$.
Product of the roots of equation $\left( i \right)$, is equal to $\dfrac{c}{a}$.
As we know roots of quadratic equation are: $\alpha $ and $\beta $.
Sum of roots:
\[\alpha + \beta = - \dfrac{b}{a}\]
By substituting the values of $\alpha $ and $\beta $, we get
$ \Rightarrow \left( { - 2} \right) + \left( { - 3} \right) = - \dfrac{b}{a}$
$ \Rightarrow - 2 - 3 = - \dfrac{b}{a}$
On addition of negative terms, we get
$ \Rightarrow - 5 = - \dfrac{b}{a}$
$ \Rightarrow \dfrac{b}{a} = 5$
Product of roots:
$\alpha .\beta = \dfrac{c}{a}$
By substituting the values of $\alpha $ and $\beta $, we get
$ \Rightarrow \left( { - 2} \right)\left( { - 3} \right) = \dfrac{c}{a}$
On multiplication of terms, we get
$ \Rightarrow 6 = \dfrac{c}{a}$
$ \Leftrightarrow \dfrac{c}{a} = 6$
Let us substitute the value in equation $\left( {ii} \right)$
${x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0......\left( {ii} \right)$
$ \Rightarrow {x^2} + 5x + 6 = 0$
Hence, ${x^2} + 5x + 6 = 0$ is an quadratic equation with roots $ - 2$ and $ - 3$.

Note:
While taking the product of two numbers always keep in mind that the product of two negative numbers is positive. Remember that the sum of roots formula itself has a negative sign. If you miss that you will get $\dfrac{b}{a}$ as $ - 5$. Then the whole equation you’ll present will be wrong.