Question

a) x =$\sqrt{24-2x}$
b) $\sqrt{6+5x} = x$
c)x+ 6= $\sqrt{12+2x}$
d)3-x=$\sqrt{3x+1}$
e)$\sqrt{x ^{2}-3x }= x-3$
f)$\sqrt{10x-6x ^{2} } =3x-1$

Answer 1

a) $x= \sqrt{24-2x}$  ⇒ $x^{2}$ = 24-2x⇒$x^{2}$+2x-24 = 0⇒ Δ=4+96=100
$x_{1}= \frac{-b+ \sqrt{delta} }{2a} = \frac{-2+10}{2} = \frac{8}{2}=4$
$x_{2}= \frac{-b- \sqrt{delta} }{2a}= \frac{-2-10}{2}= \frac{-12}{2}=-6$
b) $\sqrt{6+5x}=x$ ⇒6+5x=x²⇒x²-5x-6=0 ⇒ Δ=25+24 =69⇒$x_{1}= \frac{5+7}{2}= \frac{12}{2}=6 x_{2}= \frac{5-7}{2}= \frac{-2}{2}=-1$
c) x+6=$\sqrt{12+2x}$ ⇒ x²+12x+36 = 12+2x ⇒ x²+10x+24 = 0 ⇒Δ= 100-96 = 4 ⇒[tex] x_{1}= \frac{-10+2}{2} = \frac{-8}{2}= -4
 x_{2}= \frac{-10-2}{2}= \frac{-12}{2}=-6 [/tex]
d) 3-x = $\sqrt{3x+1}$ ⇒ 9-6x+x²=3x+1 ⇒x²-9x+8 = 0 ⇒ Δ=81-32 = 49
x₁=[tex] \frac{9+7}{2}= \frac{16}{2} =8
 x_{2}= \frac{9-7}{2}= 1 [/tex]
e) $\sqrt{ x^{2} -3x} = x-3$⇒ x²-3x = x²-6x+9 ⇒ 3x-9 = 0 ⇒3x = 9 ⇒x=$\frac{1}{3}$
f) $\sqrt{10x-6 x^{2} }$=3x-1 ⇒10x-6x²= 9x²-6x+1 ⇒ 15x²-16x+1 =0 ⇒ Δ= 256-60= 196[tex] x_{1}= \frac{16+14}{30}= \frac{30}{30}=1
 x_{2} = \frac{16-14}{30}= \frac{2}{30}= \frac{1}{15} [/tex]