Question

1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(1) t²-15​

Answer 1

AnswEr:-

Zeroes of the polynomial = √15 & - √15

Step-by-step explanation:

☯Given:-

• Polynomial = t² - 15

☯To find:-

• Zeroes of the polynomial = ?
• Verify the relationship between the zeroes and the coefficients.

☯Solution:-

Givdn polynomial:-

• t² - 15

Therefore,

⇒ t² - 15 = 0

⇒ t² = 15

⇒ t = $\pm$ √15

So,the zeroes are +√15 & -√15

$\therefore{\underline{\sf{Zeroes \: of \: polynomial = + \sqrt{15} \: \& \: -\sqrt{15} }}}$

$\rule{200}{2}$

Here,zeroes are +√15 & -√15.

Therefore,

• $\alpha$ = + √15
• $\beta$ = - √15

And the given polynomial:-

• t² - 15 = 0

Here,

• a = 1
• b = 0
• c = -15

Verifying the relationships between the zeroes & coefficients:-

First relationship :-

⇒ Sum of zeroes = -b/a

⇒ +√15 + (-√15) = -0/1

⇒ √15 - √15 = 0

⇒ 0 = 0 [Verified!]

$\rule{100}1$

2nd relationship:-

⇒ Product of zeroes = c/a

⇒ √15 × (-√15) = -15/1

⇒ -15 = -15 [Verified!]

Answer 2

Answer:

We Have Given Polynomial : t² – 15

$:\implies\tt f(t) = 0\\\\\\:\implies\tt t^2 - 15 = 0\\\\\\:\implies\tt t^2 = 15\\\\\\:\implies\tt t = \pm\:\sqrt{15}\\\\\\:\implies\underline{\boxed{\tt t = \sqrt{15} \quad or \quad - \:\sqrt{15}}}$

$\rule{160}{2}$

$\underline{\bigstar\:\textsf{Relation b/w zeroes and coefficient :}}$

$\longrightarrow\:\:\sf t^2-15=0\\\\ \longrightarrow \: \:\sf a = 1 \quad b = 0 \quad c = -\:15\\\\\\ \qquad\bf{\dag}\:\: \underline\textsf{Sum of Zeroes :} \\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{- \:b}{a}\\\\\\\dashrightarrow\tt\:\: \sqrt{15} +( - \: \sqrt{15}) = \frac{ - \:0}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt0 =0}}} \\\\\\\\\qquad\bf{\dag}\:\: \underline\textsf{Product of Zeroes :}\\\dashrightarrow\tt\:\: \alpha \beta =\dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \sqrt{15} \times( - \: \sqrt{15}) = \dfrac{ - \:15}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt - \:15 = - \:15}}}$