
If the roots of the equation \[{x^2} + px + q = 0\;\] differ by$1$, then
(A) ${p^2} = 4q$
(B) ${p^2} = 4q + 1$
(C) ${p^2} = 4q - 1$
(D) None of these
Answer
219.3k+ views
Hint: In this question, we are given that the roots of the quadratic equation \[{x^2} + px + q = 0\;\] differ by$1$. We have to calculate the sum and the product of the roots, i.e., $\alpha + \beta = \dfrac{{ - B}}{A}$, $\alpha \beta = \dfrac{C}{A}$ where the equation is \[A{x^2} + Bx + C = 0\]. Then apply the algebraic identity ${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$ and put in the required values.
Formula Used:
Quadratic equations in general: \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha$ and $\beta$
Therefore, the sum of roots is $\alpha + \beta = \dfrac{{ - B}}{A}$ and the product of roots is $\alpha \beta = \dfrac{C}{A}$
The algebraic identity is${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Complete step by step Solution:
Given that, \[{x^2} + px + q = 0\;\], is a quadratic equation.
Let, the roots of the given equation be $\alpha$ and $\beta$
According to the question, the roots of the equation differ by$1$.
It implies that $\alpha - \beta = 1$ ------- (1)
Now, compare the equation \[{x^2} + px + q = 0\;\]with the general quadratic equation, i.e., \[A{x^2} + Bx + C = 0\].
We obtain$A = 1$,$B = p$, and$C = q$.
Using the formula for sum and the product of the roots, the sum of the roots,
\[\alpha + \beta = \dfrac{{ - B}}{A}\]
It implies that,
\[\alpha + \beta = - p\] --------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
$\Rightarrow \alpha \beta = q$ --------(3)
Now, applying the algebraic identity ${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Substituting the required values in the above equation, (from equations (1), (2), and (3)),
It will be ${\left( { - p} \right)^2} = {\left( 1 \right)^2} + 4q$
On solving, we get ${p^2} = 4q+ 1$
Hence, the correct option is (B).
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 variables $\left( x \right)$ 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.
Formula Used:
Quadratic equations in general: \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha$ and $\beta$
Therefore, the sum of roots is $\alpha + \beta = \dfrac{{ - B}}{A}$ and the product of roots is $\alpha \beta = \dfrac{C}{A}$
The algebraic identity is${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Complete step by step Solution:
Given that, \[{x^2} + px + q = 0\;\], is a quadratic equation.
Let, the roots of the given equation be $\alpha$ and $\beta$
According to the question, the roots of the equation differ by$1$.
It implies that $\alpha - \beta = 1$ ------- (1)
Now, compare the equation \[{x^2} + px + q = 0\;\]with the general quadratic equation, i.e., \[A{x^2} + Bx + C = 0\].
We obtain$A = 1$,$B = p$, and$C = q$.
Using the formula for sum and the product of the roots, the sum of the roots,
\[\alpha + \beta = \dfrac{{ - B}}{A}\]
It implies that,
\[\alpha + \beta = - p\] --------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
$\Rightarrow \alpha \beta = q$ --------(3)
Now, applying the algebraic identity ${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Substituting the required values in the above equation, (from equations (1), (2), and (3)),
It will be ${\left( { - p} \right)^2} = {\left( 1 \right)^2} + 4q$
On solving, we get ${p^2} = 4q+ 1$
Hence, the correct option is (B).
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 variables $\left( x \right)$ 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.
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