Question

How do you write the quadratic equation whose roots are 1 and -5 and whose leading coefficient is 3?

Answer 1

Factoring reduces the higher degree equation into its linear equation. Quadratic equation is reduced into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term. In the above given question, we have been given the roots of the equation and the coefficient so we need to form the quadratic equation using the formula.

Complete step by step solution:
$a{x^2} + bx + c$ is a general way of writing quadratic equations where a, b and c are real rational numbers.

But here we do not know the value of a, b and c.

So, since we have been given roots in the question so let $\alpha$and $\beta$ be the roots of a quadratic equation, then the formula of quadratic equation is ${x^2} - (\alpha + \beta )x + \alpha \beta$ where $\alpha + \beta$ is the sum of the roots and $\alpha \beta$ is the product of the roots.

Therefore, Whenever we solve the standard quadratic equation, we get to know that b is the sum of the roots and c is the product of the roots.

Therefore,$\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$

Therefore, by substituting the value of $\alpha$and $\beta$,we get

$$\Rightarrow {x^2} - (1 + ( - 5))x + (1)( - 5) = 0 \\\ \Rightarrow {x^2} + 4x - 5 = 0 \\\$$

Hence, we get the above quadratic equation.

Note: In quadratic equation, an alternative way of solving the above question is by reversing the last step to the first step. So, starting with the last step it can be written as $(x - 1)(x - 5) = 0$.

Then multiplying with given coefficient 3, we get

$$\Rightarrow 3(x - 1)(x - 5) = 0 \\\ \Rightarrow 3{x^2} + 12x - 15 = 0 \\\ \Rightarrow {x^2} + 4x - 5 = 0 \\\$$