Question

if x=
$1 \div 2 - \sqrt{2}$

then, find the value of
${x}^{3} - 2 {x}^{2} - 7x + 5$





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Answer 1

Answer:

$x3−2x2−7x+5</p><p></p><p>=></p><p></p><p>= {x}^{3} - 7x - 2 {x}^{2} + 5=x3−7x−2x2+5</p><p></p><p>= x( {x}^{2} - 7) - 2 {x}^{2} + 5=x(x2−7)−2x2+5</p><p></p><p>.</p><p></p><p>Now,</p><p></p><p>put</p><p></p><p>x = \frac{1}{2} - \sqrt{3}x=21−3</p><p></p><p></p><p>by rationalisation</p><p></p><p>x = \frac{1}{2 - \sqrt{3} }x=2−31</p><p></p><p>= \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 - \sqrt{3} }=2−31×2−32+3</p><p></p><p>= 2 + \sqrt{3}=2+3</p><p></p><p>value of X then we get,</p><p></p><p>{x}^{2} = {(2 + \sqrt{3}) }^{2}x2=(2+3)2</p><p></p><p>= 4 + 3 + 4 \sqrt{3}=4+3+43</p><p></p><p>= 7 + 4 \sqrt{3}=7+43</p><p></p><p></p><p></p><p>\bold{By using above eq }Byusingaboveeq</p><p></p><p>= x( {x}^{2} - 7) - 2 {x}^{2} + 5</p><p></p><p>= (2 - \sqrt{3} )(7 + 4 \sqrt{3} - 7 ) - 2(7 + 4 \sqrt{3} ) + 5=(2−3)(7+43−7)−2(7+43)+5</p><p></p><p>= 8 \sqrt{3} + 12 - 14 - 8 \sqrt{3} + 5=83+12−14−83+5</p><p></p><p>- 2 + 5−2+5</p><p></p><p>= 3=3$

Hope it helps you

Answer 2

Question-

If  $1 \div 2-\sqrt{2}$, then find $x^3-2x^2-7x+5$.

Answer-

Explanation-

$x^3-2x^2-7x+5$

$=(1 \div 2-\sqrt{2})^3-2\times(1 \div 2-\sqrt{2})^2-7\times(1 \div 2-\sqrt{2})+5$

$=1 \div 8-\sqrt{2^3} -2\times1\div4-2-7\times1\div-\sqrt{2} +5$

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