Question

Show that the polynomials given below have no real zeroes
(i) x2 - 2x + 5
(ii) x2 + 6x + 10​

Answer 1

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Q) Show that the Polynomials given below have no real roots :

1. x² - 2x + 5
2. x² + 6x + 10

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★ Concept :

• In this question we will simply find the Discriminant of the Polynomial .

• For roots to be un real (not-real) , the Discriminant (d) must be smaller than zero .

• Standard form of a quadratic equation :

$\boxed{ \sf \: a {x}^{2} + bx + c \: }$

Where , a ≠ 0

★ Solution :

➜ First Equation -

$\rm \: {x}^{2} - 2x + 5$

here ,

• a = 1
• b = -2
• c = 5

So ,

$\rm \: discriminant = {b}^{2} - 4ac$

$\rm \: d = {( - 2)}^{2} - 4 \times 1 \times 5$

$\rm \: d = 4 - 20$

$\rm \: d = - 16$

as u can see , the d here is -16

$\rm \: d < 0$

So ,

This equation has no real roots .

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➜ Second Equation -

$\rm \: {x}^{2} + 6x + 10$

here ,

• a = 1
• b = 6
• c = 10

so ,

$\rm \: discriminant = {b}^{2} - 4ac$

$\rm \: d = {6}^{2} - 4 \times 1 \times 10$

$\rm \: d = 36 - 40$

$\rm \: d = - 4$

as u can see , here , d is -4

$\rm \: d < 0$

so ,

this equation also has no real roots .

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So , both the Polynomial have no real roots .

Answer 2

Hope this will help you..