Question

Find a quadratic equation whose sum and product of zeros are -4 and -1 respectively.

Answer 1

Hint: We will use the equations of sum and product of the roots of a quadratic equation in standard form. Then we can take a standard quadratic equation and divide it throughout with the coefficient of ${x^2}$ . Then we will make proper substitution to obtain the required quadratic equation.

Complete step-by-step answer:
We know that for a quadratic equation of the form $a{x^2} + bx + c = 0$ , the sum of their zeros is given by, $sum = \dfrac{{ - b}}{a}$ and product of the zeros are given by, product $ = \dfrac{c}{a}$ .
It is given that the sum of the roots is -4. So, we can write,
 $ \Rightarrow - 4 = \dfrac{{ - b}}{a}$
On cancelling the negative signs on both sides, we get,
 $ \Rightarrow \dfrac{b}{a} = 4$ … (1)
In the question, it is also given that the product of the roots is -1.
So, we can write,
 $ \Rightarrow \dfrac{c}{a} = - 1$ …. (2)
 Now we can take a quadratic equation in its standard form. It is given by,
 $a{x^2} + bx + c = 0$
Now we can divide the equation throughout with a.
 \[ \Rightarrow \dfrac{{a{x^2} + bx + c}}{a} = \dfrac{0}{a}\]
On splitting the numerator, we get,
 \[ \Rightarrow \dfrac{a}{a}{x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\]
On simplification, we get,
 \[ \Rightarrow {x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\]
Now we can substitute equation (1) and (2). Then we will get,
 \[ \Rightarrow {x^2} + \left( 4 \right)x + \left( { - 1} \right) = 0\]
On simplification, we get,
 \[ \Rightarrow {x^2} + 4x - 1 = 0\]
Therefore, the required quadratic equation is \[{x^2} + 4x - 1 = 0\] .

Note: Alternate solution to this problem is given by,
Let $\alpha $ and $\beta $ be the roots of the quadratic equation.
Then the sum of the roots is given by,
 $\alpha + \beta = - 4$ …. (a)
And product of the root is given by,
 $\alpha \beta = - 1$
On dividing throughout with $\alpha $ , we get,
 $ \Rightarrow \beta = \dfrac{{ - 1}}{\alpha }$ … (b)
Now we can substitute (b) in (a). So, we will get,
 $ \Rightarrow \alpha - \dfrac{1}{\alpha } = - 4$
On multiplying throughout with $\alpha $ , we get
 $ \Rightarrow {\alpha ^2} - 1 = - 4\alpha $
 $ \Rightarrow {\alpha ^2} + 4\alpha - 1 = 0$
As $\alpha $ is the root of the equation, we can write the equation in terms of x.
 $ \Rightarrow {x^2} + 4x - 1 = 0$
Therefore, the required quadratic equation is \[{x^2} + 4x - 1 = 0\] .