Question

if alpha and beta are zeros of the polynomial 3 X square + 5 x + 13 then find​

Answer 1

Step-by-step explanation:

plz complete your question......it doesn't mention what to find......

Answer 2

Step-by-step explanation:

I think your question is to find the roots I. e. alpha and beta.

$let \: f(x) = 3 {x}^{2} + 5x + 13$

Now there are two ways to do this. one is by trying special values of x. but it turns to be a cumbersome process, so I would set the polynomial=0 and solve the quadratic equation by applying the well sridhar acharya's formula

here we get, a=3, b=5, c=13

so

$x = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \\ or \\ x = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

let's first evaluate the common term

$\sqrt{ {b}^{2} - 4ac } \\ = \sqrt{ {5}^{2} - (4 \times 3 \times 13)} \\ = \sqrt{25 - 156} \\ = \sqrt{ - 131} \\ = \sqrt{131}i$

so,

$x = \frac{ - 5 + \sqrt{131}i}{6} \\ or \\ x = \frac{ - 5 - \sqrt{131}i}{6}$

so we get

$\alpha = \frac{ - 5 + \sqrt{131}i}{6} \\ or \\ \beta = \frac{ - 5 - \sqrt{131}i}{6}$

hope fully you know what is refered to as i. If not then,

$i = \sqrt{ - 1}$

which is an imaginary number.

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