Question

What are zeroes of quadratic eqn 9x² + 6x + 1​

Answer 1

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To find zeroes

p(x)=0

9x² + 6x + 1 = 0

9x² + 3x + 3x + 1 = 0

3x(3x+1)+1(3x+1)=0

(3x+1)(3x+1)=0

(3x+1)²=0

3x+1=0

3x= -1

Thus, the quadratic eqn has equal roots

$x = \frac{ - 1}{3}$

Answer 2

Answer:

Step-by-step explanation:

Hello,

We have,

9x²+6x+1

There are 3 ways with which we can find zeroes of this equation,factorization,using quadratic formula or by completing the square method...

1.facforisation,

9x²+6x+1

Splitting the middle term

9x²+3x+3x+1

3x(3x+1)+1(3x+1)

(3x+1)(3x+1)

Hence we get 2 factors therefore,zeroes are -1/3,-1/3

2.By using quadratic formula

X=-b+_root b²-4ac/2a

X=-6+_root 6²-9×4×1/2×9

X=-6+_0/18

X=-6/18=-1/3,-1/3

3.By completing the square method,

9x²+6x+1

X²+2/3×+1/9=0

X²+2/3X=-1/9

X²+2/3X+(1/3)²=-1/9+(1/3)²

(X+1/3)²=-1/9+1/9

(X+1/3)²=0

X+1/3=0

X=-1/3,-1/3

Hope it helps,please mark me as brainliest......