Question

√2x^2-2x-√3=0 the root of the equation

Answer 1

Answer:

x=\frac{\sqrt{2}\left(-\sqrt{\sqrt{6}+1}+1\right)}{2}

x=

2

2

(−

6

+1

+1)

Step-by-step explanation:

2

x

2

−2x−

3

=0

ax^{2}+bx+c=0

\frac{-b±\sqrt{b^{2}-4ac}}{2a}

±

\sqrt{2}x^{2}-2x-\sqrt{3}=0

2

x

2

−2x−

3

=0

ax^{2}+bx+c=0

\sqrt{2}

a-2b-\sqrt{3}

c\frac{-b±\sqrt{b^{2}-4ac}}{2a}

x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\sqrt{2}\left(-\sqrt{3}\right)}}{2\sqrt{2}}

x=

2

2

−(−2)±

(−2)

2

−4

2

(−

3

)

-2

x=\frac{-\left(-2\right)±\sqrt{4-4\sqrt{2}\left(-\sqrt{3}\right)}}{2\sqrt{2}}

x=

2

2

−(−2)±

4−4

2

(−

3

)

-4\sqrt{2}

x=\frac{-\left(-2\right)±\sqrt{4+\left(-4\sqrt{2}\right)\left(-\sqrt{3}\right)}}{2\sqrt{2}}

x=

2

2

−(−2)±

4+(−4

2

)(−

3

)

-4\sqrt{2}

-\sqrt{3}

x=\frac{-\left(-2\right)±\sqrt{4+4\sqrt{6}}}{2\sqrt{2}}

x=

2

2

−(−2)±

4+4

6

44\sqrt{6}

x=\frac{-\left(-2\right)±\sqrt{4\sqrt{6}+4}}{2\sqrt{2}}

x=

2

2

−(−2)±

4

6

+4

4+4\sqrt{6}

x=\frac{-\left(-2\right)±2\sqrt{\sqrt{6}+1}}{2\sqrt{2}}

x=

2

2

−(−2)±2

6

+1

-22

x=\frac{2±2\sqrt{\sqrt{6}+1}}{2\sqrt{2}}

x=

2

2

2±2

6

+1

x=\frac{2±2\sqrt{\sqrt{6}+1}}{2\sqrt{2}}

±22\sqrt{1+\sqrt{6}}

x=\frac{2\sqrt{\sqrt{6}+1}+2}{2\sqrt{2}}

x=

2

2

2

6

+1

+2

2+2\sqrt{1+\sqrt{6}}

2\sqrt{2}

x=\frac{\sqrt{2}\left(\sqrt{\sqrt{6}+1}+1\right)}{2}

x=

2

2

(

6

+1

+1)

x=\frac{2±2\sqrt{\sqrt{6}+1}}{2\sqrt{2}}

±2\sqrt{1+\sqrt{6}}

2

x=\frac{-2\sqrt{\sqrt{6}+1}+2}{2\sqrt{2}}

x=

2

2

−2

6

+1

+2

2-2\sqrt{1+\sqrt{6}}

2\sqrt{2}

x=\frac{\sqrt{2}\left(-\sqrt{\sqrt{6}+1}+1\right)}{2}

x=

2

2

(−

6

+1

+1)

x=\frac{\sqrt{2}\left(\sqrt{\sqrt{6}+1}+1\right)}{2}

x=

2

2

(

6

+1

+1)