Question

find the zeroes of the given polynomial 9x^2-5 . please write the solution in step​

Answer 1

Since the polynomial has no linear terms which means that the zeroes will be the same number but the other one will be negative and it's a quadratic polynomial so it'll have at Max 2 zeroes. Hope it helps you.

Answer 2

GivEn:-

• 9x² - 5 = 0

To find:-

• Zeroes of the polynomial.

Solution:-

★ 9x² - 5 = 0

$:\implies$ 9x² = 5

$:\implies$ x² = $\dfrac{5}{9}$

$:\implies$ x = ± $\sqrt{ \dfrac{5}{9}}$

$:\implies$ x = $\sqrt{ \dfrac{5}{9}}$ , $- \sqrt{ \dfrac{5}{9}}$

$\rule{200}3$

★ Verification:-

Let $\sf \alpha = \sqrt{ \dfrac{9}{5}}$ , $\sf \beta = - \sqrt{ \dfrac{9}{5}}$

We know that,

Sum of zeroes = $\sf ( \alpha + \beta ) = \dfrac{- b}{a}$

$:\implies\sf \bigg( \cancel{ \sqrt{ \dfrac{5}{9}}} \cancel{ - \sqrt{ \dfrac{5}{9}}} \bigg) = - \dfrac{0}{9}$

$:\implies\sf 0 = 0$

$:\implies\sf LHS = RHS$

Now,

Product of zeroes = $\sf ( \alpha \beta ) = \dfrac{c}{a}$

$:\implies\sf \bigg( \sqrt{ \dfrac{5}{9}} \bigg) \times \bigg( - \sqrt{ \dfrac{5}{9}} \bigg) = - \dfrac{- 5}{9}$

$:\implies\sf \dfrac{- 5}{9} = - \dfrac{- 5}{9}$

$:\implies\sf LHS = RHS$

★ Hence Verified!!

$\rule{200}3$