Answer

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**Hint:**In this problem, we have to find the quadratic equation. As we know, a quadratic equation gives two roots of the given variable. In this question, we will solve the given two conditions of the roots and thus put those values in the quadratic equation, to get which is the solution for the answer.

**Complete step by step answer:**

According to the question, we have to find the quadratic equation.

Thus, we will use the formula of the roots, to get the required solution for the problem.

As we know, the general form of the quadratic equation is $a{{x}^{2}}+bx+c=0$ and its roots are $\alpha \text{ }and\text{ }\beta $ .

Thus, we will first solve the two given statements in the problem, which is

$\alpha +\beta =3,$ ------- (1)

$\alpha -\beta =4$ -------- (2)

Now, we will find the value of $\alpha $ from equation (1) that is we will subtract $\beta $ from both sides in the equation (1), we get

$\alpha +\beta -\beta =3-\beta $

As we know, the same terms with opposite signs cancel out each other, therefore we get

$\alpha =3-\beta $ ------- (3)

Now, we will use the Substitution method that is we will substitute the value of the above equation in equation (2), we get

$3-\beta -\beta =4$

On further solving, we get

\[3-2\beta =4\]

Now, we will subtract 3 on both sides in the above equation, we get

$3-2\beta -3=4-3$

As we know, the same terms with opposite signs cancel out each other, therefore we get

$-2\beta =1$

Now, we will divide -2 on both sides in the above equation, we get

$\dfrac{-2}{-2}\beta =\dfrac{1}{-2}$

On further simplification, we get

$\beta =\dfrac{-1}{2}$

Thus, we will substitute the above value of the equation in equation (3), we get

$\alpha =3-\left( \dfrac{-1}{2} \right)$

On further simplification, we get

$\alpha =3+\left( \dfrac{1}{2} \right)$

Now, we will take the LCM of the above equation, we get

$\alpha =\dfrac{6+1}{2}$

Therefore, we get

$\alpha =\dfrac{7}{2}$

Thus, we get the values of the roots, that is $\alpha =\dfrac{7}{2}$ and $\beta =\dfrac{-1}{2}$ .

Now, we will substitute the above values of the roots in the given quadratic equations one by one, and if we get the value equal to 0, it implies it is the root of the quadratic equation.

So we will first use option (A) of the given problem, which is

$A)4{{x}^{2}}-12x-7=0$ ------ (4)

Now, we will first substitute the value $\alpha =\dfrac{7}{2}$ on the LHS in equation (4), we get

$4{{\left( \dfrac{7}{2} \right)}^{2}}-12.\dfrac{7}{2}-7$

On further solving, we get

$4\left( \dfrac{49}{4} \right)-12.\dfrac{7}{2}-7$

Therefore, we get

$\begin{align}

& 49-42-7 \\

& 49-49 \\

\end{align}$

$0=RHS$

Therefore, $\alpha =\dfrac{7}{2}$ is the root of the given quadratic equation.

Now, we will first substitute the value $\beta =\dfrac{-1}{2}$ on the LHS in the equation (4) , we get

$4{{\left( \dfrac{-1}{2} \right)}^{2}}-12.\dfrac{(-1)}{2}-7$

On further solving, we get

$4\left( \dfrac{1}{4} \right)+12.\dfrac{1}{2}-7$

Therefore, we get

$\begin{align}

& 1+6-7 \\

& 7-7 \\

\end{align}$

$0=RHS$

Therefore, $\beta =\dfrac{-1}{2}$ is the root of the given quadratic equation.

Thus, both $\alpha $ and $\beta $ are the roots of the equation $4{{x}^{2}}-12x-7=0$ .

So we will solve option (B) of the given problem, which is

$B)4{{x}^{2}}-12x+7=0$------ (5)

Now, we will first substitute the value $\alpha =\dfrac{7}{2}$ on the LHS in the equation (5), we get

$4{{\left( \dfrac{7}{2} \right)}^{2}}-12.\dfrac{7}{2}+7$

On further solving, we get

$4\left( \dfrac{49}{4} \right)-12.\dfrac{7}{2}+7$

Therefore, we get

$\begin{align}

& 49-42+7 \\

& 56-42 \\

\end{align}$

$14\ne RHS$

Therefore, $\alpha =\dfrac{7}{2}$ is not the root of the given quadratic equation.

Now, we will first substitute the value $\beta =\dfrac{-1}{2}$ on the LHS in the equation (5) , we get

$4{{\left( \dfrac{-1}{2} \right)}^{2}}-12.\dfrac{(-1)}{2}+7$

On further solving, we get

$4\left( \dfrac{1}{4} \right)+12.\dfrac{1}{2}+7$

Therefore, we get

$\begin{align}

& 1+6+7 \\

& 7+7 \\

\end{align}$

$14\ne RHS$

Therefore, $\beta =\dfrac{-1}{2}$ is not the root of the given quadratic equation.

So we will solve option (C) of the given problem, which is

$C)4{{x}^{2}}-12x-25=0$ ------ (6)

Now, we will first substitute the value $\alpha =\dfrac{7}{2}$ on the LHS in the equation (6), we get

$4{{\left( \dfrac{7}{2} \right)}^{2}}-12.\dfrac{7}{2}-25$

On further solving, we get

$4\left( \dfrac{49}{4} \right)-12.\dfrac{7}{2}-25$

Therefore, we get

$\begin{align}

& 49-42-25 \\

& 56-67 \\

\end{align}$

$-11\ne RHS$

Therefore, $\alpha =\dfrac{7}{2}$ is not the root of the given quadratic equation.

Now, we will first substitute the value $\beta =\dfrac{-1}{2}$ on the LHS in the equation (6) , we get

$4{{\left( \dfrac{-1}{2} \right)}^{2}}-12.\dfrac{(-1)}{2}-25$

On further solving, we get

$4\left( \dfrac{1}{4} \right)+12.\dfrac{1}{2}-25$

Therefore, we get

$\begin{align}

& 1+6-25 \\

& 7-25 \\

\end{align}$

$-18\ne RHS$

Therefore, $\beta =\dfrac{-1}{2}$ is not the root of the given quadratic equation.

Thus, for the two real numbers $\alpha $ and $\beta $ are such that $\alpha +\beta =3,\alpha -\beta =4$ , then $\alpha $ and $\beta $ are the roots of the quadratic equation $4{{x}^{2}}-12x-7=0$ .

**So, the correct answer is “Option A”.**

**Note:**While solving this problem, do mention the formulas you are using to avoid an error. One of the alternative methods to solve this problem is first to find the roots of each quadratic equation and then find the sum and product of roots and use them to find the quadratic equation.

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