Without solving, examine the nature of roots of the equation.

4 $x^{2}$ – 4x + 1 = 0
[2 Marks]

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Solution

Given equation is,

$4 x^{2} - 4 x + 1 = 0$

$\Rightarrow a = 4, b = - 4, c = 1$
[1 Mark]

Value of discriminant is

$D = b^{2} - 4 a c$

$\Rightarrow D = (- 4)^{2} - 4 (4) (1)$

$\Rightarrow D = 16 - 16 = 0$

Since D = 0,

Roots of quadratic equation are real and equal.
[1 Mark]

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