Question

Detemine the nature of roots of the quadratic equation x2-6x+9=0

Answer 1

Given ,

• The polynomial is x² - 6x + 9

Here ,

a = 1

b = -6

c = 9

We know that , the discriminant of polynomial is given by

$\boxed{ \sf{D = {(b)}^{2} - 4ac }}$

Thus ,

D = (-6)² - 4 × 1 × 9

D = 36 - 36

D = 0

Therefore ,

• The nature of roots are real and equal

Answer 2

The roots of the quadratic equation x² - 6x + 9 = 0 are real and equal

Given :

The quadratic equation x² - 6x + 9 = 0

To find :

The nature of the roots of the quadratic equation x² - 6x + 9 = 0

Concept :

General form of a quadratic equation is

ax² + bx + c = 0

The Discriminant of the quadratic equation is denoted by D and defined as

D = b² - 4ac

Case : I When D ≥ 0 then roots are real

Case : II When D = 0 then roots are equal and real

Case : III When D < 0 then roots are unequal and imaginary.

Case : IV When D > 0 and a perfect square then the roots are real, rational and unequal.

Case : V When D > 0 and not a perfect square then the roots are real, irrational and unequal.

Solution :

Step 1 of 3 :

Write down the quadratic equation

Here the given Quadratic equation is

x² - 6x + 9 = 0

Step 2 of 3 :

Find discriminant of the equation

Comparing the given Quadratic equation x² - 6x + 9 = 0 with the general equation ax² + bx + c = 0 we get

a = 1 , b = - 6 , c = 9

The discriminant of the quadratic equation

= b² - 4ac

$\sf = {( - 6)}^{2} - 4 \times 1 \times 9$

$\sf = 36 - 36$

$\sf = 0$

Step 3 of 3 :

Find nature of the roots of the quadratic equation

Since the discriminant of the quadratic equation = D = 0

Hence the roots of the quadratic equation x² - 6x + 9 = 0 are real and equal

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