Question

find the zeros of the polynomial x^2 -2x+3​

Answer 1

Hello bro,

Your answer with step by step explanation is mentioned below.

Step-by-step explanation:

So let's start : )

To find zero(s) of any polynomial, we simply let it 0,

And then solve the value of the variable.

So first we will lets the given equation be 0

${x}^{2} - 2x + 3 = 0$

Before solving, I would recommend you to see is the zeros of polynomial imaginary or not.

You can test it by calculating.

If b²>4ac, we get real zeros.

If b²<4ac, we get imaginary zeros.

Here,

b= -2, a = 1, c=3

So clearly, as (-2)²<4×1×3, this polynomial has imaginary roots. (Zeros are also known as roots)

Interesting right : )

So lets move on and solve for it.

If a polynomial has imaginary roots, we can only solve it using the quadratic formula which is,

$\frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}$

where,

a is coefficient of x²

, b is coefficient of x

& c is constant

This mean there are two zeros,

So first zero=

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

Substituting the value of a, b & c i.e. 1, -2, 3 respectively, we get,

$\frac{- ( - 2) + \sqrt{ {( - 2)}^{2} -4 \times 1 \times 3 } }{2 \times 1}$

Evaluating the above expression, we get,

$\frac{2+ \sqrt{4 - 12} }{2}$

$\frac{2 + \sqrt{ - 8} }{2}$

$\frac{2}{2} + \frac{ \sqrt{ - 8} }{2}$

$2 + \frac{2 \sqrt{ - 2} }{2}$

$1 + \sqrt{ - 2}$

, and thats the first zero

Similarly,

Substituting the value of a,b and c on this expression,

$\frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

, and then evaluating, we get

$1 - \sqrt{ - 2}$

, and thats the second zero.

So, in brief the zeros are:

$1 + \sqrt{ - 2} \: \: \: \: \: and \: \: \: \: \: 1 - \sqrt{ - 2}$

Hope it helps,

And,

Don't forget to follow me and choose this answer as the brainliest one.