Question

Verify weather the following are zeroes of the polynomial, indicated against them.
(i) $p(x) = 3x {}^{2} - 1, \: x = - \frac{1}{ \sqrt{3} } ,\frac{2}{ \sqrt{3} }$

(ii) p(x) = 5x - π, x = $\frac{4}{5}$

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Answer 1

Step-by-step explanation:

(ii) p(x) = 5x – π, x = 4/5

Plug x = 4/5 we get

=> p(4/5) = 5x – π,

=> p(4/5) = 5(4/5) – π,

=> p(4/5) = 4 – π,

And pi = 22/7 so that

=> p(4/5) = 4 – 22/7 is not = 0

Answer 2

Answer:

$\huge{\underline{\textsf{\textbf{~~~Question}}}}$

Verify weather the following are zeroes of the polynomial, indicated against them.

(i) $p(x) = 3x {}^{2} - 1, \: x = - \frac{1}{ \sqrt{3} } ,\frac{2}{ \sqrt{3} }$

(ii) p(x) = 5x - π, x = $\frac{4}{5}$

$\huge{\underline{\textsf{\textbf{~~~Solution}}}}$

(i) $p(x) = 3x {}^{2} - 1, \: x = - \frac{1}{ \sqrt{3} } ,\frac{2}{ \sqrt{3} }$

given :-

• $\sf p(x) = 3x {}^{2} - 1$ is a equation

• $\ x = \frac{1}{ \sqrt{3} } ,\frac{2}{ \sqrt{3} }$

putting the value of x in equation

$p(x) = 3( \frac{1}{ \sqrt{3} } ) {}^{2} - 1 = 0\\ → 3( \frac{1}{3}) - = 0 \\ → \frac{ \cancel3}{ \cancel3} - 1 = 0 \\ →1 - 1 = 0 \\ \sf \color{teal} { † \: hence \: verified} \\ p(x) = 3( \frac{2}{ \sqrt{3} } ) {}^{2} - 1 = 0\\ → 3( \frac{2}{3}) - = 0 \\ →\frac{4}{3} - 1 = 0 \\ →\frac{4 - 3}{3} = 0 \\ → \frac{1}{3} ≠ 0 \\ \sf \color{teal} { † \: hence \: not\: verified}$

(ii) p(x) = 5x - π, x = $\frac{4}{5}$

given :-

• p(x) = 5x - π is a equation

• x = $\frac{4}{5}$

$p(x) = 5x - \pi \\ = 5( \frac{4}{5} ) - \pi = 0\\ = \frac{20}{5} - \pi = 0 \\ = 4 - \pi = 0\\ \sf \: we \: know \: that \: value \: of \: \pi \: is \: \frac{22}{7} \\ = 4 - \frac{22}{7} \\ = \frac{28 - 22}{7} = 0 \\ = 6≠0 \\ \sf \: hence \: not \: verifided$

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$\huge{\color{purple}{\textsf{\textbf{Purple~you♡}}}}$