Question

23. Show that the polynomial x2 – 2x + 5 has no real zeroes.​

Answer 1

Step-by-step explanation:

Using the quadratic equation :

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

Let the modified equation be :

${x}^{2}+( - 2x)+5$

Let the values of 'a', 'b' and 'c' be :

a = 1

b = -2

c = 5

Inserting the values in the equation,

$x = \frac{ - ( - 2) + - \sqrt{ { (- 2)}^{2} - 4 \times 1 \times 5 } }{2 \times 1}$

$x = \frac{ 2 + - \sqrt{ {4} - 20 } }{2}$

$x = \frac{2 + - \sqrt{ - 16} }{2}$

Since there is a negative sign under a square root, the roots of the equation are not possible.

Hence proved.

(P.S. - Please mark as Brainliest if you are satisfied with my answer.)

Answer 2

Step-by-step explanation:

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