Question

Find the zeroes of quadratic polynomial h(t)=t^2-15

Answer 1

t^2-15= 0

t^2 = 15

t = √15

.the zeros are √15 , -√15Mark as a Brainlist answer

Answer 2

AnsWer :

t = √15 , -√15.

Solution :

When, h(t) = t² -15.

• then, t²- 15 = 0.

Let try with Alzebric method,

$\tt\dashrightarrow {t}^{2} - 15 = 0.$

$\tt\dashrightarrow {t}^{2} = 15$

$\tt\dashrightarrow t = \pm \sqrt{15} .$

$\rule{200}1$

Let try with Quadratic formula,

We have equation, t²-15.

Compare with General Equation.

$\tt a {x}^{2} + bx + c = 0. \: \: \: \red{a \neq 0.}$

• a = 1.
• b = 0.
• c = -15.

$\tt x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Putting the value, which given above, We get.

$\tt\dashrightarrow t = \frac{0 \pm \sqrt{ {(0)}^{2} - 4 \times 1 \times - 15 } }{2 \times 1}$

$\tt\dashrightarrow t = \frac{ \sqrt{60} }{2}$

$\begin{array}{r | l} 2 & 60 \\ \cline{2-2} 3 & 30 \\ \cline{2-2} 5 & 10 \\ \cline{2-2} 2 & 2 \\ \cline{2-2} & 1 \end{array}$

$\tt\dashrightarrow t = \frac{ \cancel{2} \sqrt{15} }{\cancel{2}}$

$\tt\dashrightarrow t = \pm \sqrt{15}.$

Therefore, the value of the equation be √15 and -√15.