Math, asked by Creepyboy95, 3 months ago

Find zeros of polynomial

 \\  \sqrt[5]{5} {x}^{2}  + 30x +  \sqrt[8]{5}
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Answers

Answered by VinCus
28

Required Answer:-

 \\\longrightarrow \tt \:  \sqrt[5]{5} {x}^{2}  + 30x +  \sqrt[8]{5}

➡️By splitting the middle term,

\\ \longrightarrow \tt \:  \sqrt[5]{5} {x}^{2}  + 20x + 10x +  \sqrt[8]{5}

\\ \longrightarrow \tt \:  \sqrt[5]{5} {x}^{2}  + (5  \times  4)x + (5 \times 2)x +  \sqrt[8]{5}

\\ \longrightarrow \tt \:  \sqrt[5]{5} {x}^{2}  +  (\sqrt{5}  \times  \sqrt{5}  \times 4)x + (5 \times 2)x +  \sqrt[8]{5}

\\ \longrightarrow \tt \:  \sqrt[5]{x}  \times (5x +  \sqrt[4]{5}) + 2(5x +  \sqrt[4]{5})

\\ \longrightarrow \tt \: (5x +  \sqrt[4]{5}) \: ( \sqrt{5}x + 2)

\\ \longrightarrow \tt \: (5x +  \sqrt[4]{5}) \implies(1) \\

➡️From (1)

\\ \longrightarrow \tt \: 5x  =  - \sqrt[4]{5}

 \\  \longrightarrow \tt \: x  =   \frac{ -  \sqrt[4]{5} }{5}

 \longrightarrow \tt \: ( \sqrt{5}x + 2) \implies \: (2)\\

➡️From (2)

\\ \longrightarrow \tt \:  \sqrt{5}x + 2  = 0

\\ \longrightarrow \tt \:  \sqrt{5}x  = - 2

 \\  \longrightarrow \tt \:  x  =  \frac{ - 2}{ \sqrt{5} }

➡️The two Zeroes of the polynomial is :-

 \\  \longrightarrow \tt \: \frac{ -  \sqrt[4]{5} }{5} ,  \frac{ - 2}{ \sqrt{5} }

Answered by gurmanpreet1023
23

Answer:

Zeroes of the polynomial are\frac{  - 2\sqrt{5} }{5}, \frac{  - 4\sqrt{5} }{5}

Step-by-step explanation:

Given equation,

5 root 5 x square + 30x + 8 root 5

This can be written as,

5√5x²+ 30 x + 8√5

We will factorise the given equation by splitting the middle term method

5√5 x² + 20 x + 10 x + 8√5

10 x can be written as ( 5 * 2) x

5√5 x² + 20 x + (5*2) x + 8√5

Also, 5 can be written as √5 * √5

So now the Equation becomes :

= 5√5 x² + 20 x+ (√5 * √5 * 2)x+ 8√5 = 5x ( √5 x + 4 ) + √5 * 2 ( √5 x+ 4)

= 5x ( √5 x + 4) + 25 ( √5 x + 4) = ( 5x + 2√5 ) (√5 x + 4 )

Zeroes are

5x + 2√5 =0

5x = -2√5

 \frac{ - 2 \sqrt{5} }{5}

Also,

√5 x + 4 = 0

√5 x = - 4

x =  \frac{ - 4}{ \sqrt{5} }

Multiplying and dividing by √5

x =  \frac{ - 4}{ \sqrt{5} }  \times  \frac{ \sqrt{5} }{ \sqrt{5} }

x =  \frac{ - 4 \sqrt{5} }{5}

Hence,

Zeroes of the polynomial are -2v5, -15


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