Question

$2 \sqrt{2x {}^{2} } + 11x + 6 \sqrt{2}$
answer this question please​

Answer 1

Solve

$2 \sqrt{2} {x}^{2} + 11x + 6 \sqrt{2}$

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Solution..

$= > 2 \sqrt{2} {x}^{2} + 11x + 6 \sqrt{2} \\ \\ = > 2 \sqrt{2} {x}^{2} + 8x + 3x + 6 \sqrt{2} \\ \\ = > 2 \sqrt{2} x(x + 2 \sqrt{2} ) + 3(x + 2 \sqrt{2} ) \\ \\ = > (2 \sqrt{2} x + 3) \: \: \: (x + 2 \sqrt{2} ) \\ \\ = > 2 \sqrt{2} x + 3 = 0 \\ = > 2 \sqrt{2} x = - 3 \\ \color{red} \boxed{= > x = \frac{ - 3}{2 \sqrt{2} }} \\ \\ \\ = > x + 2 \sqrt{2} = 0 \\ \color{red} \boxed{ = > x = - 2 \sqrt{2} }$

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So the two zeroes or factors of 2√2x²+11x+6√2 ....are

$</u></strong><strong><u>\</u></strong><strong><u>c</u></strong><strong><u>o</u></strong><strong><u>l</u></strong><strong><u>o</u></strong><strong><u>r</u></strong><strong><u>{</u></strong><strong><u>orange</u></strong><strong><u>}</u></strong><strong><u>\huge{ \frac{ - 3}{2 \sqrt{2} } \: \: \: \: \: \: or \: \: \: \: \: \: - 2 \sqrt{2} }$

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Rules to find factors or zeroes

(1) First multiply 1st term with last term

i.e 2√2x² x 6√2. = 24

(2) Then split middle term in that form so that the product of two number will be 24x² and Subtraction/addition will be 11x

(3) After this, find common factors between 1st two terms and then other two

i.e refers above .

(4) After find common factors ,Put both factors equal to zero and find zeroes or factors....

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Hope it's helps you ✨

Answer 2

Answer:

Solve

2 \sqrt{2} {x}^{2} + 11x + 6 \sqrt{2}22x2+11x+62

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Solution..

\begin{gathered} = > 2 \sqrt{2} {x}^{2} + 11x + 6 \sqrt{2} \\ \\ = > 2 \sqrt{2} {x}^{2} + 8x + 3x + 6 \sqrt{2} \\ \\ = > 2 \sqrt{2} x(x + 2 \sqrt{2} ) + 3(x + 2 \sqrt{2} ) \\ \\ = > (2 \sqrt{2} x + 3) \: \: \: (x + 2 \sqrt{2} ) \\ \\ = > 2 \sqrt{2} x + 3 = 0 \\ = > 2 \sqrt{2} x = - 3 \\ \color{red} \boxed{= > x = \frac{ - 3}{2 \sqrt{2} }} \\ \\ \\ = > x + 2 \sqrt{2} = 0 \\ \color{red} \boxed{ = > x = - 2 \sqrt{2} }\end{gathered}=>22x2+11x+62=>22x2+8x+3x+62=>22x(x+22)+3(x+22)=>(22x+3)(x+22)=>22x+3=0=>22x=−3=>x=22−3=>x+22=0=>x=−22

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So the two zeroes or factors of 2√2x²+11x+6√2 ....are

< /u > < /strong > < strong > < u > \ < /u > < /strong > < strong > < u > c < /u > < /strong > < strong > < u > o < /u > < /strong > < strong > < u > l < /u > < /strong > < strong > < u > o < /u > < /strong > < strong > < u > r < /u > < /strong > < strong > < u > { < /u > < /strong > < strong > < u > orange < /u > < /strong > < strong > < u > } < /u > < /strong > < strong > < u > \huge{ \frac{ - 3}{2 \sqrt{2} } \: \: \: \: \: \: or \: \: \: \: \: \: - 2 \sqrt{2} }</u></strong><strong><u> </u></strong><strong><u>c</u></strong><strong><u>o</u></strong><strong><u>l</u></strong><strong><u>o</u></strong><strong><u>r</u></strong><strong><u></u></strong><strong><u>orange</u></strong><strong><u></u></strong><strong><u>22−3or−22

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Rules to find factors or zeroes

(1) First multiply 1st term with last term

i.e 2√2x² x 6√2. = 24

(2) Then split middle term in that form so that the product of two number will be 24x² and Subtraction/addition will be 11x

(3) After this, find common factors between 1st two terms and then other two

i.e refers above .

(4) After find common factors ,Put both factors equal to zero and find zeroes or factors....

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Hope it's helps you ✨