Question

4s²-4s+1 find zeroes.​

Answer 1

Step-by-step explanation:

hi friend 

here the quadratic equation is 4s²-4s+1_______________________(1)

By using factorization method

4s²-4s+1=0

4s²-2s-2s+1=0

2s(2s-1)-1(2s-1)=0

(2s-1)(2s-1)=0

either (2s-1) is its zero or (2s-1) = 0

so the zeros are s=\frac{1}{2}21 ,s=\frac{1}{2}21 

put the value of s in eq (1)

we got 4 X \frac{1}{4} -4X \frac{1}{2} +1 = 1 - 2+1=2-2=04X41−4X21+1=1−2+1=2−2=0 

hence the factor lead equation becomes zero so,

hence verified

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