Respuesta :
[tex]\\ \sf\longmapsto \sqrt{(x^2-2x-2)}+\sqrt{2x-3}=5[/tex]
We know
if x=√a+√b then
- x^2=a+b+2√ab
[tex]\\ \sf\longmapsto x^2-2x-2+2x-3+2\sqrt{(x^2-2x-2)(2x-3)}=5^2[/tex]
[tex]\\ \sf\longmapsto x^2-3+2\sqrt{2x^3-4x^2-4x-3x^2+6x+6}=25[/tex]
[tex]\\ \sf\longmapsto 2\sqrt{2x^3-7x^2+2x+6}=25-x^2+3=28-x^2[/tex]
[tex]\\ \sf\longmapsto \sqrt{2x^3-7x^2+2x+6}=\dfrac{28-x^2}{2}[/tex]
[tex]\\ \sf\longmapsto 2x^3-7x^2+2x+6=\dfrac{(28-x^2)^2}{4}[/tex]
[tex]\\ \sf\longmapsto 2x^3-7x^2+2x+6=\dfrac{784-56x^2+x^4}{4}[/tex]
[tex]\\ \sf\longmapsto 8x^3-28x^2+8x+24=784-56x^2+x^4[/tex]
[tex]\\ \sf\longmapsto x^4-8x^3-28x^2-8x+760=0[/tex]
- Solving further
The equation has no real solutions .(attached a pic of calculator )
Answer:
[tex]{ \rm{ \sqrt{( {x}^{2} - 2x - 2) } + \sqrt{(2x - 3)} = 5 }}[/tex]
• let's first collect like terms
[tex]{ \rm{ \sqrt{( {x}^{2} - 2x - 2)} = 5 - \sqrt{(2x - 3)} }}[/tex]
• Then, let's take a square:
[tex]{ \rm{ {( \sqrt{( {x}^{2} - 2x - 2)}) }^{2} = {(5 - \sqrt{(2x - 3)} )}^{2} }} \\ \\ { \rm{ {x}^{2} - 2x - 2 = (5 - {(2x - 3)}^{ \frac{1}{2} } )(5 - {(2x - 3)}^{ \frac{1}{2} }) }} \\ \\ { \rm{ {x}^{2} - 2x - 2 = 25 - 10 \sqrt{2x - 3} - 2x - 3 }} \\ \\ { \rm{ {x}^{2} - 2x + 2x - 2 - 22 + 10 \sqrt{2x - 3} = 0}} \\ \\ { \rm{ {x}^{2} + 10 \sqrt{2x - 3} - 22 = 0}} \\ \\ { \rm{the \: roots \: of \: the \: equation \: are \: not \: real}}[/tex]
Answer: x = 3(-1 + i) or 3(-1 - i)
