How do you solve #5x ^ { 2} - 2= 33#?

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Answer 1 · 2017-12-18T03:17:13

#:. x=pmsqrt7#

#5x^2-2=33#

We want the RH side to be 0, so subtract 33 from both sides:

#5x^2-2color(red)(-33)=33color(red)(-33)#

#5x^2-35=0#

Both terms on the LH side are divisible by 5 (and we can divide by 5 on the RH side), so let's divide by 5:

#(5x^2-35)/color(red)(5)=0/color(red)(5)#

#x^2-7=0#

#x^2=7#

#:. x=pmsqrt7#

Graphically, we can see this if we graph the RH and LH sides:

graph{(y-5x^2+2)(y-0x-33)=0[-6,6,30,40]}