Question

Hi any one help me

If x=1/2 (√3+1) than the value of 4x^3+2x^2-8x+2

Answer 1

Solution of the question is given below in the image

Answer 2

Hey friend,
Here is the answer you were looking for:
$x = \frac{1}{2( \sqrt{3} + 1) } \\ \\ = \frac{1}{2 \sqrt{3} + 2} \times \frac{2 \sqrt{3} - 2}{2 \sqrt{3} - 2} \\ \\ = \frac{2 \sqrt{3} - 2}{ {(2 \sqrt{3}) }^{2} - {(2)}^{2} } \\ \\ = \frac{2 \sqrt{3} - 2}{12 - 4} \\ \\ = \frac{2 \sqrt{3} - 2 }{8} \\ \\ = \frac{ \sqrt{3} - 1}{4} \\ \\ 4 {x}^{3} + 2 {x}^{2} - 8x + 2 \\ \\ = 4 {( \frac{ \sqrt{3} - 1}{4} )}^{3} + 2 {( \frac{ \sqrt{3} - 1 }{4} )}^{2} - 8( \frac{ \sqrt{3} - 1}{4} ) + 2 \\ \\ = 4( \frac{ \sqrt{3} }{4} - \frac{1}{4} )^{3} + 2( \frac{ \sqrt{3} }{4} - \frac{1}{4} )^{2} - 2 \sqrt{3} + 2 + 2 \\ \\ = 4( {( \frac{ \sqrt{3} }{4}) }^{3} - {( \frac{1}{4} )}^{3} - 3( \frac{ \sqrt{3} }{4} )( \frac{4}{4} )( \frac{ \sqrt{3} }{4} - \frac{1}{4} )) + 2(( { \frac{ \sqrt{3} }{4}) }^{2} + {( \frac{1}{4}) }^{2} + 2( \frac{ \sqrt{3} }{4} )( \frac{1}{4} )) - 2 \sqrt{3} + 4 \\ \\ = 4(( \frac{3 \sqrt{3} }{64} - \frac{1}{64} - \frac{12 \sqrt{3} }{16} \times \frac{ \sqrt{3} }{4} - \frac{12 \sqrt{3} }{16} \times \frac{1}{4} ) + 2( \frac{3}{16} + \frac{1}{16} + \frac{2 \sqrt{3} }{16} ) - 2 \sqrt{3 } + 4 \\ \\ = 4( \frac{3 \sqrt{3} }{64} - \frac{1}{64} - \frac{9 }{16} - \frac{3 \sqrt{3} }{16} ) + 2( \frac{1}{4} + \frac{2 \sqrt{3} }{16} ) - 2 \sqrt{3} + 4 \\ \\ = 4 \times \frac{3 \sqrt{3} }{64} - 4 \times \frac{1}{64} - 4 \times \frac{9}{16} - 4 \times \frac{3 \sqrt{3} }{16} + 2 \times \frac{1}{4} + 2 \times \frac{2 \sqrt{3} }{16} - 2 \sqrt{3} + 4 \\ \\ = \frac{3 \sqrt{3} }{16} - \frac{1}{16} - \frac{9}{4} - \frac{3 \sqrt{3} }{4} + \frac{1}{2} + \frac{ \sqrt{3} }{4 } - 2 \sqrt{3} + 4 \\ \\ = \frac{3 \sqrt{3} - 1 - 9 \times 4 - 3 \sqrt{3} \times 4 + 1 \times 8 + \sqrt{3} \times 4 - 2 \sqrt{3} \times 16 + 4 \times 16}{16} \\ \\ = \frac{3 \sqrt{3} - 1 - 36 - 12 \sqrt{3} + 8 + 4 \sqrt{3} - 32 \sqrt{3} + 64}{16} \\ \\ = \frac{ - 37 \sqrt{3} + 35}{16} \\ \\ = \frac{35 - 37 \sqrt{3} }{16}$


Hope this helps!!!

Thanks...
☺