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

A quadratic equation is written as $ax^2+bx+c$ $\,=\,$ $0$ in standard form and it has two solutions, which are called the zero or roots of a quadratic equation.

The sum by adding the two roots of a quadratic equation is called the sum of roots of a quadratic equation. Let’s denote the two zeros of a quadratic equation by two Greek alphabets $\alpha$ and $\beta$.

$\alpha$ $\,=\,$ $\dfrac{-b+\sqrt{b^2\,-\,4ac}}{2a}$

$\beta$ $\,=\,$ $\dfrac{-b\,-\sqrt{b^2\,-\,4ac}}{2a}$

Now, add the two roots of a quadratic equation to find the sum of them mathematically.

$\implies$ $\alpha+\beta$ $\,=\,$ $\dfrac{-b+\sqrt{b^2\,-\,4ac}}{2a}$ $+$ $\dfrac{-b\,-\sqrt{b^2\,-\,4ac}}{2a}$

The expression on the right hand side of the equation expresses the addition of two fractions. Now, let’s add them by the addition rule of the fractions.

$\,\,=\,\,$ $\dfrac{-b+\sqrt{b^2\,-\,4ac}+\big(-b-\sqrt{b^2\,-\,4ac}\big)}{2a}$

$\,\,=\,\,$ $\dfrac{-b+\sqrt{b^2\,-\,4ac}-b-\sqrt{b^2\,-\,4ac}}{2a}$

$\,\,=\,\,$ $\dfrac{-b-b+\sqrt{b^2\,-\,4ac}-\sqrt{b^2\,-\,4ac}}{2a}$

$\,\,=\,\,$ $\dfrac{-2b+\cancel{\sqrt{b^2\,-\,4ac}}-\cancel{\sqrt{b^2\,-\,4ac}}}{2a}$

$\,\,=\,\,$ $\dfrac{-2b}{2a}$

$\,\,=\,\,$ $-\dfrac{2b}{2a}$

$\,\,=\,\,$ $-\dfrac{2 \times b}{2 \times a}$

$\,\,=\,\,$ $-\dfrac{\cancel{2} \times b}{\cancel{2} \times a}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\alpha+\beta$ $\,=\,$ $-$ $\dfrac{b}{a}$

It is derived that the sum of two roots of a quadratic equation is called the negative quotient of the coefficient of $x$ divided by the coefficient of $x$ square.

$4x^2+5x+6 \,=\, 0$

Let’s learn how to find the sum of the zeros of a quadratic equation. Now, compare the quadratic equation with the standard form a quadratic equation. Now, the values of $a = 4$, $b = 5$ and $c = 6$.

$\,\,\,\therefore\,\,\,\,\,\,$ $-\dfrac{b}{a} \,=\, -\dfrac{5}{4}$

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