Question

$\sqrt{2x^2-5x+2}$-$\sqrt{x^2-x+2}$=$\sqrt{x^2-3x+2}$

Answer 1

$\sqrt{ 2x^{2}-5x+2 }- \sqrt{ x^{2} -x+2}= \sqrt{ x^{2} -3x+2}$
 ridicam ambele parti la patrat:
[tex] 2 x^{2} -5x+2-2 \sqrt{(2x^2-5x+2)( x^{2} -x+2}+ \sqrt{ x^{2} -x+2}= x²-3x+2
[tex] -2\sqrt{ (2x^{2}-5x+2 )(x^{2} -x+2)}= x^2-3x+2-2 x^{2} +5x-2 [/tex]
$-2\sqrt{ (2x^{2}-5x+2 )(x^{2} -x+2)}=- x^{2} +2x$
[tex]( -2\sqrt{ (2x^{2}-5x+2 )(x^{2} -x+2)})^2=(- x^{2} +2x)^2 [/tex]
[tex]4(2x^{2}-5x+2 )(x^{2} -x+2)}=-x^4-3x+2 [/tex]
$8x^4-8x^3+16x^2-20x^3+20x^2-40x+8x^2-8x+16+x^4+4x^3+$+4x²=0
$9x^4-24x^3+48x^2-48x+16=0$ vom imparti ecuatia la x² si vom obtine:
9x²-24x+48-$\frac{48}{x}+ \frac{16}{x^2}=0 sau 9(x^2+ \frac{1}{x^2}-24(x- \frac{1}{x})+48=0$
Notam: $x+ \frac{1}{x}=t si x^{2} + \frac{1}{ x^{2} }= t^{2} -2$  dupa care obtinem ecuatia: 9(t²-2)-24t+8=0
9t²-24t+30=0 impartim la 3 si vom obtine
3t²-8t+10=0
a=3  b=-8  c=10
Δ=b²-4ac=64-120=-56
$t_{1,2} \frac{8+- i \sqrt{56} }{6}= \frac{4+-i \sqrt{14} }{3}$
x₁,₂,₃,₄=$\frac{4+-i \sqrt{14} }{3}$ ---acestea sunt radacinile ecuatiei de gradul IV