Question

Find the zeroes of quadratic polynomial h(t) =t2 —3 and prove the eelation between the zeroes and the coefficients.

Answer 1

SolutioN :

We have,

$\tt \dagger \: \: \: \: \: h(t) = {t}^{2} - 3.$

☛ Condition :

• t Zero of h( t ) = t² - 3.

$\tt : \implies {t}^{2} - 3 = 0.$

$\tt : \implies {t}^{2} = 3.$

$\tt : \implies t = \pm \sqrt{ 3}.$

$\rule{200}3$

Let's try with Other method :

$\tt \dagger \: \: \: \: \: \fbox{ x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

Where as,

• a = 1.
• b = 0.
• c = - 3.

$\tt :\implies x = \dfrac{ 0\pm \sqrt{ 0 - 4 \times 1 \times - 3 } }{2}$

$\tt :\implies x = \dfrac{ \pm \sqrt{ 12} }{2}$

$\tt :\implies x = \dfrac{2 \pm \sqrt{ 3} }{2}$

$\tt :\implies x = \pm \sqrt{3} .$

Let's, Zero be

• α = √3.
• β = - √3.

✎ Sum of Zeros.

→ α + β = √3 - √3.

→ - b / a = 0.

→ 0 = 0.

$\rule{90}1$

✎ Product Of Zero.

→ α * β = √3 * - √3.

→ α * β = - 3.

→ c / a = - 3.

→ - 3 = - 3.

✡ Hence Verify