Question

positive real root of 81x^2-1=0​

Answer 1

$81 {x}^{2} - 1 = 0$

$81 {x}^{2} = 1$

${x}^{2} = \frac{1}{81}$

$x = \sqrt{ \frac{1}{81} }$

$x = \frac{1}{9} \: \: and \: - \frac{1}{9}$

positive real root is

$x = \frac{1}{9}$

Answer 2

Positive real root is $\bf \red{x = \frac{1}{9}}$

Given:

• $81 {x}^{2} - 1 = 0$

To find:

• Find the positive real root.

Solution:

Identity to be used:

$\bf ( {a}^{2} - {b}^{2} ) = (a + b)(a - b)$

Step 1:

Rewrite the equation.

$81 {x}^{2} - 1 = ( {9x)}^{2} - ( {1)}^{2}$

here,

$\bf a = 9x$

and

$\bf b = 1$

so,

$(9x - 1)(9x + 1) = 0$

Step 2:

Equate both factors with zero.

$9x - 1 = 0$

or

$\bf x = \frac{1}{9}$

or

$9x + 1 = 0$

or

$x = - \frac{1}{9}$

Thus,

Positive real root is $\bf x = \frac{1}{9}$

Learn more:

1) (ii) If p{x) = 2x2 + 4x + 6 is a quadratic polynomial then what is the value of sum of zeroes?

a) 0

b) 1

c) - 2

d) 4

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2) find the zero of the polynomial p(x) =3x+5

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