Question

Find the zeros of the following quadratic polynomial and verify the
relationship between zeros and the coefficient
t2-15

Answer 1

$\texttt{\textsf{\large{\underline{Solution!}}}}$

Given:

→ t² - 15 = 0

→ (t)² - (√15)² = 0

Using identity a² - b² = (a + b)(a - b), we get:

→ (t + √15)(t - √15) = 0

By zero product rule:

→ (t + √15) = 0 or (t - √15) = 0

→ t = √15, -√15.

⊕ So, the zeros of the given equation are √15 and -√15.

$\texttt{\textsf{\large{\underline{Verification!}}}}$

Comparing the given equation with ax² + bx + c = 0, we get:

→ a = 1

→ b = 0

→ c = -15

Here, sum of zeros = √15 - √15 = 0

Also, sum of zeros = -b/a = 0/1 = 0  (Verified)

Again, product of zeros = -√15 ×  √15 = -15

Also, product of zeros = c/a = -15/1 = -15 (Verified)

Answer 2

Given :-

• t² - 15

now :-

• by using the identity (a² - b²) = (a - b) (a + b)

by using the identity we can write it as :-

• t² - 15 = (t - √15) (X + √15)

so therefore the value of t² - 15 is zero

when t = √15 or t = -√15

verification :-

sum of zeros = a + b √15 + (-√15) = 0

$:\implies\sf \: - \frac{coefficient \: of \: t}{coefficient \: of \: t ^{2} } = - \frac{0}{1} = 0$

now by using product of zero = ab = (√15) (-√15)

$:\implies\sf \: \frac{constant \: term \: }{cofficient \: of \: {t}^{2} } = \frac{15}{1} = - 15$

hence our answer is verified..