How do you solve $x - 6 = \sqrt{6 x + 4}$ $x- 6= \sqrt { 6x + 4}$ ?

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Answer 1 · 2017-05-11T10:24:46

$x - 6 = \sqrt{6 x + 4}$ $x- 6= \sqrt { 6x + 4}$

Taking squares in both sides

${(x - 6)}^{2} = 6 x + 4$ $(x-6)^2=6x+4$

$x^{2} - 12 x + 36 - 6 x - 4 = 0$ $x^2-12x+36-6x-4=0$

$x^{2} - 18 x + 32 = 0$ $x^2-18x+32=0$

$x^{2} - 16 x - 2 x + 32 = 0$ $x^2-16x-2x+32=0$

$(x - 16) \cdot (x - 2) = 0$ $(x-16)*(x-2)=0$

Because $x - 6 = \sqrt{6 x + 4} \geq 0 \Rightarrow x \geq 6$ $x-6=sqrt(6x+4)>=0=>x>=6$

Hence the only acceptable solution is $x = 16$ $x=16$

How do you solve x−6=√6x+4x- 6= \sqrt { 6x + 4}?