Question

$\sf \: \alpha \: and \: \beta \: are \: the \: root \: of \: the \: equation \: \\ \sf {2y}^{2} + 6y - 5 = 0 \: then \: find \: the \: value \: \\ \sf \: of \: \alpha + \beta$



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Answer 1

Given: A quadratic equation, $2y^2 + 6y - 5$, and its roots as α and β.

To find: The value of α+β.

Solution: First, we find the roots of the given equation,

Since this is a quadratic equation in y, it must have two roots.

Using the formula $D= b^2 - 4ac$, where D is the determinant, and a, b and c are coefficients of $y^2, y$ and the constant respectively, we check the nature of the roots.

From given equation,

• a=2
• b=6
• c=-5

Plugging these values in the discriminant equation, we get the answer as D= 76.

As D>0, the roots are real and distinctive.

The value of α+β is given by the formula $\alpha +\beta = \frac{-b}{a}$.

Plugging in the values of a and b, we get α+β= -3.

Hence the required solution is α+β= -3.

Answer 2

$\sf \fbox \red{question}$

$\sf \: = {2y}^{2} + 6y - 5 = 0$

$\sf \fbox \red{soluction}$

$\sf \: y = \frac{ - 6± \sqrt{ {6}^{2} - 4(2)( - 5)} }{2(2)}$

$\sf \: y = \frac{ - 6± \sqrt{36 + 40} }{4}$

$\sf \: = \frac{ - 6± \sqrt{76} }{4}$

$\sf \: = \frac{ - 6± \sqrt{4 \times 19} }{4}$

$\sf \: = \frac{ - 6± \sqrt[2]{19} }{4}$

$\sf \: = \frac{ - 3± \sqrt{19} }{2}$

$\sf \: = here. \: \alpha = \frac{ - 3 + \sqrt{19} }{2} . \beta = \frac{ - 3 - \sqrt{19} }{2}$

$\sf \: = \alpha + \beta = \frac{ - 3 + \sqrt{19 } - 3 - \sqrt{19} }{2}$

$\sf \: = \frac{ - 6}{2}$

$\sf \: = - 3$

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

$\sf \fbox \red{ans = α \: + \: β = - 3}$