Question

If the roots of ${x^2} - bx + c = 0$ are two consecutive integers then ${b^2} - 4c$ is

Answer 1

Hint: In this question, we are given two consecutive integers are the roots of the equation ${x^2} - bx + c = 0$. We have to find the value of ${b^2} - 4c$. Assume the roots be $\alpha $ and $\alpha + 1$. Calculate the sum and product of the root i.e., $\alpha + \beta = \dfrac{{ - B}}{A}$, $\alpha \beta = \dfrac{C}{A}$ where the equation is \[A{x^2} + Bx + C = 0\]. Using the equation of the sum of the roots calculate the value of roots and put the required values in the product.

Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Given that,
Two consecutive integers are the roots of ${x^2} - bx + c = 0$
Let, the roots of the equation be $\alpha $ and $\alpha + 1$
Compare the equation ${x^2} - bx + c = 0$ with the general quadratic equation \[A{x^2} + Bx + C = 0\]
We get, $A = 1$, $B = - b$, and $C = c$
Now, using the formula of sum and product of the root
Sum of the roots, \[\alpha + \alpha + 1 = \dfrac{{ - B}}{A}\]
It implies that, \[\alpha + \alpha + 1 = b\]
$2\alpha + 1 = b$
$2\alpha = b - 1$
On solving, we get \[\alpha = \dfrac{1}{2}\left( {b - 1} \right)\] ------(1)
Adding $1$ on both sides, we get
 \[\alpha + 1 = \dfrac{1}{2}\left( {b - 1} \right) + 1\]
\[\alpha +1 = \dfrac{1}{2}\left( {b + 1} \right)\]--------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
\[\alpha \left( {\alpha + 1} \right) = c\]
Put \[\alpha = \dfrac{1}{2}\left( {b - 1} \right)\] and \[\alpha + 1 = \dfrac{1}{2}\left( {b + 1} \right)\] in the above equation (from equation (1) and (2))
It implies that,
\[\dfrac{1}{2}\left( {b - 1} \right) \times \dfrac{1}{2}\left( {b + 1} \right) = c\]
Also, written as $\left( {b - 1} \right)\left( {b + 1} \right) = 4c$
Using the algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
We get, ${b^2} - 1 = 4c$
Or ${b^2} - 4c = 1$
Hence, if the roots of the equation ${x^2} - bx + c = 0$ are two consecutive integers the ${b^2} - 4c$ is equal to $1$.

Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.