Question

the roots of the quadratic equation 16x square-8x+1=0​

Answer 1

Answer:

16x²-8x+1=0

16x²-4x-4x+1=0

4x(4x-1) -(4x-1) =0

(4x-1) (4x-1) =0

(4x-1) ²=0

4x-1=0

x=1/4, 1/4

Answer 2

$Given \: Quadratic \: equation :\\ 16x^{2}-8x+1=0$

$\implies (4x)^{2} - 2 \times (4x) \times 1 + 1^{2}=0$

$\implies ( 4x - 1 )^{2} = 0$

/* By Algebraic Identity */

$\boxed { \pink { a^{2} - 2ab + b^{2} = (a-b)^{2} }}$

$\implies (4x-1)(4x-1) = 0$

$\implies 4x-1 = 0\: Or \:4x-1= 0$

$\implies 4x = 1 \: Or \:4x = 1$

$\implies x = \frac{1}{4} \: Or \:x = \frac{1}{4}$

Therefore.,

$\green { x = \frac{1}{4} \: Or \:x = \frac{1}{4}} \\\green{ are \: two \: equal \:roots \:of \:the \: equation}$

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