Question

Find zeroes of polynomial and verify it
$5 {x}^{2} - 2$
​

Answer 1

Given:

☛ polynomial p(x) = 5x² - 2

To Find:

☛ Zeroes of p(x)

Also,

Verify it.

Solution:

☛ p(x) = 5x² - 2

➜ 5x² - 2 = 0

➜ 5x² = 2

➜ x = ± √2/5

Zeroes are √2/5 and -√2/5

Proof:

If √2/5 and -√2/5 are Zeroes of p(x) then they must satisfy the polynomial p(x):

x = √2/5

☛ 5x² - 2

➜ 5(√2/5)² - 2

➜ 5 × 2 / 5 - 2

➜ 2 - 2

➜ 0

x = -√2/5

☛ 5x² - 2

➜ 5(-√2/5)² - 2

➜ 5 × 2/5 - 2

➜ 2 - 2

➜ 0

Hence, Proved.

Answer 2

Question:

Find the zeroes of polynomial and verify it

$5 {x}^{2} - 2$

Solution:

$5 {x}^{2} - 2 = 0 \\ \\ 5 {x}^{2} = 2 \\ \\ {x}^{2} = \frac{2}{5} \\ \\ x = + \sqrt{ \frac{2}{5} } \\ \\ and \\ \\ x = - \sqrt{ \frac{2}{5} }$

now,

we have the polynomial

$5 {x}^{2} + 0.x - 2 = 0 \\ \\ so \\ \\ a = 5 \\ b = 0 \\ c = - 2$

so,

sum of zeroes = -b/a

$\sqrt{ \frac{2}{5} } + ( - \sqrt{ \frac{2}{5} }) = \frac{ - 0}{5} \\ \\ 0 = 0 \\ \\ verified$

also,

product of zeroes =c/a

$\sqrt{ \frac{2}{5} } \times - \sqrt{ \frac{2 }{5} } = \frac{ - 2}{5} \\ \\ \frac{ - 2}{5} = \frac{ - 2}{5} \\ \\ verified$