How do you solve $\sqrt { 2x + 41} - 3= x$ ?

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Answer 1 · 2018-03-14T12:09:36

$x=-8,4$

We have:

$x=sqrt(2x+41)-3$

$x+3=sqrt(2x+41)$

$(x+3)^2=2x+41$

$x^2+6x+9=2x+41$

$x^2+4x-32=0$

We can use many methods to solve this. Let's use the quadratic formula. According to it:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Here, $a=1, b=4, c=-32$ . Inputting:

$x=(-4+-sqrt(4^2-(4)(1)(-32)))/(2*1)$

$x=(-4+-sqrt(16+126))/2$

$x=(-4+-12)/2$

$x=-2+-6$

$x=-8,4$

How do you solve \sqrt { 2x + 41} - 3= x\sqrt { 2x + 41} - 3= x?