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Find the nature of the roots of the following quadratic equation. If the real roots exist, find them: 2x2 - 6x + 3 = 0 - Mathematics

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Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:

2x2 - 6x + 3 = 0

Determine the nature of the roots of the following quadratic equation:

2x2 - 6x + 3 = 0

Sum
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Solution

2x2 - 6x + 3 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = 2, b = -6, c = 3

Discriminant = b2 - 4ac

= (-6)2 - 4 (2) (3)

= 36 - 24

= 12

As b2 - 4ac > 0,

Therefore, distinct real roots exist for this equation

x = `(-b+-b^2-4ac)/(2a)`

= `(-(-6)+-sqrt((-6)^2-4(2)(3)))/(2(2))`

= `(6+-sqrt12)/4`

= `(6+-2sqrt3)/4`

= `(3+-sqrt3)/2`

Therefore, the root are `x=(3+sqrt3)/2 and x = (3-sqrt3)/2`

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Chapter 4: Quadratic Equations - Exercise 4.4 [Page 91]

APPEARS IN

NCERT Mathematics [English] Class 10
Chapter 4 Quadratic Equations
Exercise 4.4 | Q 1.3 | Page 91
RD Sharma Mathematics [English] Class 10
Chapter 4 Quadratic Equations
Exercise 4.6 | Q 1.2 | Page 41

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