Form a quadratic equation whose sum and the product of the roots are \[ - 3\] & \[2\], respectively.
Answer
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Hint: There are different kinds of equations; they are linear equations, quadratic equations, and polynomial equations. Linear equations will have one root and the quadratic equation will have2 roots.
The general form of a quadratic equation can be written as \[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\], where \[\alpha \& \beta \] are the roots of the equation.
Complete step-by-step solution:
It is given that the sum of the roots of a quadratic equation is \[ - 3\] and the product of the roots of the quadratic equation is \[2\].
Let the roots of the required quadratic equation be \[\alpha \] and \[\beta \]. It is given that the sum of the roots is \[ - 3\] thus, \[\alpha + \beta = - 3\] and the product of the roots is \[2\] thus \[\alpha \beta = 2\].
We know that the general form of a quadratic equation is \[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\], where \[\alpha \& \beta \] are the roots of the equation.
Let us substitute the values of sum and the product of the roots.
\[ {x^2} - (\alpha + \beta )x + \alpha \beta = 0 \\
\Rightarrow {x^2} - ( - 3)x + 2 = 0 \]
On simplifying the above equation, we get
\[ \Rightarrow {x^2} + 3x + 2 = 0\]
Thus, this is the required quadratic equation whose sum of their root is \[ - 3\] and product of their root is \[2\]
Let us verify the equation whether it is correct or not.
First, let us solve the quadratic equation, \[{x^2} + 3x + 2 = 0\]
\[ \Rightarrow (x + 2)(x + 1) = 0\]
On simplifying this we get \[x = - 2\] and \[x = - 1\].
Let \[\alpha = - 2\] & \[\beta = - 1\]. Now let us verify the sum and product of the roots.
\[\alpha + \beta = \left( { - 2} \right) + \left( { - 1} \right) = - 3\]
\[\alpha \beta = \left( { - 2} \right)\left( { - 1} \right) = 2\]
Thus, we got the same sum and product of roots that is given in the problem. Hence, the quadratic equation we found is correct.
Note: If we need to find the equation when the roots are given, we first need to find what type of equation is that. Then we need to simplify the given roots then substitute them in the respective general equation. Simplifying that will give us the required equation.
The general form of a quadratic equation can be written as \[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\], where \[\alpha \& \beta \] are the roots of the equation.
Complete step-by-step solution:
It is given that the sum of the roots of a quadratic equation is \[ - 3\] and the product of the roots of the quadratic equation is \[2\].
Let the roots of the required quadratic equation be \[\alpha \] and \[\beta \]. It is given that the sum of the roots is \[ - 3\] thus, \[\alpha + \beta = - 3\] and the product of the roots is \[2\] thus \[\alpha \beta = 2\].
We know that the general form of a quadratic equation is \[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\], where \[\alpha \& \beta \] are the roots of the equation.
Let us substitute the values of sum and the product of the roots.
\[ {x^2} - (\alpha + \beta )x + \alpha \beta = 0 \\
\Rightarrow {x^2} - ( - 3)x + 2 = 0 \]
On simplifying the above equation, we get
\[ \Rightarrow {x^2} + 3x + 2 = 0\]
Thus, this is the required quadratic equation whose sum of their root is \[ - 3\] and product of their root is \[2\]
Let us verify the equation whether it is correct or not.
First, let us solve the quadratic equation, \[{x^2} + 3x + 2 = 0\]
\[ \Rightarrow (x + 2)(x + 1) = 0\]
On simplifying this we get \[x = - 2\] and \[x = - 1\].
Let \[\alpha = - 2\] & \[\beta = - 1\]. Now let us verify the sum and product of the roots.
\[\alpha + \beta = \left( { - 2} \right) + \left( { - 1} \right) = - 3\]
\[\alpha \beta = \left( { - 2} \right)\left( { - 1} \right) = 2\]
Thus, we got the same sum and product of roots that is given in the problem. Hence, the quadratic equation we found is correct.
Note: If we need to find the equation when the roots are given, we first need to find what type of equation is that. Then we need to simplify the given roots then substitute them in the respective general equation. Simplifying that will give us the required equation.
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