Question

find zero of quadratic polynomial 9y2-6y+1​

Answer 1

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Question :

Find zero of quadratic polynomial 9y²-6y+1=0

To find :

Find the roots of the given quadratic equation :

Solution :

$\\ \implies \tt9 {y}^{2} - 6y + 1 = 0$

$\\ \implies \tt \: 9 {y}^{2} - 3y - 3y + 1 = 0$

$\\ \implies \tt \: 3y(3y - 1) - 1(3y - 1) = 0$

$\\ \implies \tt \: (3y - 1) = 0$

$\\ \implies \tt \: 3y - 1 = 0$

$\\ \implies \tt \: y = \frac{1}{3}$

➡ The zeroes of this quadratic equations is $\sf{\dfrac{1}{3}}$

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Answer 2

Answer:

this is the answer

Step-by-step explanation:

answer 1/3