Find the nature of the roots of the following quadratic equations. 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
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Solution
2x2 - 6x + 3 = 0
Comparing this 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 `(3+-sqrt3)/2 `
Concept: Nature of Roots
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