1.

If alpha and beta are the zeroes of √2x^2+3x+√2 then find the value of alpha + beta + alpha×beta​

Answer»

\dfrac{( - <klux>3</klux>+\sqrt{2}) \sqrt{2}}{2}

Step-by-step explanation:

Given:

Alpha and Beta are the zeroes of given equation.

To FIND:

Alpha + Beta + Alpha × Beta

Solution:

COMPARING the equation √2x² + 3x + √2 from ax² + bx + c, I GET,

a = √2, b = 3 and c = √2

We know that,

\text{Sum of roots} = -\frac{b}{a} \\ \\ i.e \: \: \: \alpha + \beta = -\frac{3}{ \sqrt{2} }

\text{Product of roots} = \frac{c}{a} \\ \\ i.e \: \: \: \alpha \beta = \frac{ \sqrt{2} }{ \sqrt{2} } \\ \\ = 1

Now,

\alpha + \beta + \alpha \beta \\ \\ - \frac{3}{ \sqrt{2} } + 1 \\\\\\[\text{Rationalizing the denominator}] \\ \frac{ - 3 + \sqrt{2} }{ \sqrt{2} } \\ \\ \frac{( - 3 + \sqrt{2}) \sqrt{2}}{2}

REQUIRED answer is \boxed{\blue{\dfrac{( - 3 + \sqrt{2}) \sqrt{2}}{2}}}


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