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# If $\alpha ,\beta$ are roots of the quadratic equation $2{{x}^{2}}-4x-3=0$ . Find $\alpha +\beta$ and $\alpha \beta$ .  Hint: To solve the question, we have to analyse the given quadratic equation and compare it with the general quadratic equation. The obtained values through comparison are substituted in the formula for the sum of roots of the quadratic equation and for the product of the roots of the quadratic equation, to arrive at the answer.

The given quadratic equation is $2{{x}^{2}}-4x-3=0$
Given that $\alpha ,\beta$ are the roots of the given quadratic equation.
We know that for a general quadratic equation $a{{x}^{2}}+bx+c=0$
The sum of roots of quadratic equation is equal to $\dfrac{-b}{a}$
The product of roots of quadratic equation is equal to $\dfrac{c}{a}$
On comparing our given quadratic equation with the general equation, we get
The value of coefficient of ${{x}^{2}}$ , for the general quadratic equation and for the given quadratic equation are a, 2 respectively.
Thus, a = 2
The value of coefficient of x , for the general quadratic equation and for the given quadratic equation are b, -4 respectively.
Thus, b = -4
The value of constant, for the general quadratic equation and for the given quadratic equation are c, -3 respectively.
Thus, c = -3
By substituting the above values in the formula for sum of the roots and product of the roots of the quadratic equation formula, we get
The sum of roots of the given quadratic equation is equal to $\dfrac{-(-4)}{2}=\dfrac{4}{2}=2$

The product of roots of the given quadratic equation is equal to $\dfrac{-(3)}{2}=\dfrac{-3}{2}=-1.5$

Thus, the values of $\alpha +\beta$ and $\alpha \beta$ are 2, -1.5 respectively.

Note: The possibility of mistake can be, not able to analyse and then compare the given quadratic equation with the general quadratic equation. The other possibility of mistake can be not applying the formula for the sum of roots of the quadratic equation and for the product of the roots of the quadratic equation.
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