If x = (1 + √2 + √3) Then Find the value of 2x⁴ - 8x³ - 5x² +2

1.	If x = (1 + √2 + √3) Then Find the value of 2x⁴ - 8x³ - 5x² +26x - 28.Given Options Are :1 ) √62 ) 2√33 ) 3√24 ) 6√6
Answer» ANSWER : $\underline{\bigstar\:\textsf{According to the Question Now :}}$ $:\implies\tt x = 1+\sqrt{2}+\sqrt{3}\\\\\\:\implies\tt x-1=\sqrt{2}+\sqrt{3}\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Squaring Both Sides}}\\\\:\implies\tt (x-1)^2=(\sqrt{2}+\sqrt{3})^2\\\\\\:\implies\tt x^2+1^2-(2.x.1)=(\sqrt{2})^2+(\sqrt{3})^2+(2.\sqrt{2}.\sqrt{3})\\\\\\:\implies\tt x^2+1-2x=2+3+2\sqrt{6}\\\\\\:\implies\tt x^2-2x-4=2\sqrt{6}\qquad\dfrac{\qquad}{}eq.(1)\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Squaring Both Sides}}\\\\:\implies\tt (x^2-2x-4)^2=(2\sqrt{6})^2\\$ $\\:\implies\tt (x^2)^2+(-2x)^2+(-4)^2+2[(x^2.-2x)+(-2x.-4)+(-4.x^2)]=(2.2.6)\\\\\\:\implies\tt x^4+4x^2+16+2[-2x^3+8x-4x^2]=24\\\\\\:\implies\tt x^4+4x^2+16-4x^3+16x-8x^2-24=0\\\\\\:\implies\tt x^4-4x^3-4x^2+16x-8=0\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Multiplying by 2 Both Sides}}\\\\:\implies\tt 2x^4-8x^3-8x^2+32x-16=0\\\\\\:\implies\tt2x^4-8x^3-5x^2-3x^2+26x+6x-16-12+12=0\\\\\\:\implies\tt(2x^4-8x^3-5x^2+26x-28)-3x^2+6x+12=0\\\\\\:\implies\tt2x^4-8x^3-5x^2+26x-28=3x^2-6x+12\\$ $\\:\implies\tt2x^4-8x^3-5x^2+26x-28=3(x^2-2x+4)\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Putting the value from eq.(1)}}\\\\:\implies\tt2x^4-8x^3-5x^2+26x-28=3(2\sqrt{6})\\\\\\:\implies\boxed{\tt2x^4-8x^3-5x^2+26x-28=6\sqrt{6}}$ ⠀ $\therefore\:\underline{\textsf{Required Value of the equation is \textbf{6$\sqrt{ \textbf6}$}}}$

If x = (1 + √2 + √3) Then Find the value of 2x⁴ - 8x³ - 5x² +26x - 28.Given Options Are :1 ) √62 ) 2√33 ) 3√24 ) 6√6

Answer»

ANSWER :

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\tt x = 1+\sqrt{2}+\sqrt{3}\\\\\\:\implies\tt x-1=\sqrt{2}+\sqrt{3}\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Squaring Both Sides}}\\\\:\implies\tt (x-1)^2=(\sqrt{2}+\sqrt{3})^2\\\\\\:\implies\tt x^2+1^2-(2.x.1)=(\sqrt{2})^2+(\sqrt{3})^2+(2.\sqrt{2}.\sqrt{3})\\\\\\:\implies\tt x^2+1-2x=2+3+2\sqrt{6}\\\\\\:\implies\tt x^2-2x-4=2\sqrt{6}\qquad\dfrac{\qquad}{}eq.(1)\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Squaring Both Sides}}\\\\:\implies\tt (x^2-2x-4)^2=(2\sqrt{6})^2\\$

$\\:\implies\tt (x^2)^2+(-2x)^2+(-4)^2+2[(x^2.-2x)+(-2x.-4)+(-4.x^2)]=(2.2.6)\\\\\\:\implies\tt x^4+4x^2+16+2[-2x^3+8x-4x^2]=24\\\\\\:\implies\tt x^4+4x^2+16-4x^3+16x-8x^2-24=0\\\\\\:\implies\tt x^4-4x^3-4x^2+16x-8=0\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Multiplying by 2 Both Sides}}\\\\:\implies\tt 2x^4-8x^3-8x^2+32x-16=0\\\\\\:\implies\tt2x^4-8x^3-5x^2-3x^2+26x+6x-16-12+12=0\\\\\\:\implies\tt(2x^4-8x^3-5x^2+26x-28)-3x^2+6x+12=0\\\\\\:\implies\tt2x^4-8x^3-5x^2+26x-28=3x^2-6x+12\\$

$\\:\implies\tt2x^4-8x^3-5x^2+26x-28=3(x^2-2x+4)\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Putting the value from eq.(1)}}\\\\:\implies\tt2x^4-8x^3-5x^2+26x-28=3(2\sqrt{6})\\\\\\:\implies\boxed{\tt2x^4-8x^3-5x^2+26x-28=6\sqrt{6}}$

⠀

$\therefore\:\underline{\textsf{Required Value of the equation is \textbf{6$\sqrt{ \textbf6}$}}}$