How do you find the zeros for #y=5x^2-2#?

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Answer 1 · 2018-03-20T22:59:51

You set y equal 0, then solve the resulting equation for the value(s) of x.

Given: #y=5x^2-2#

Set #y = 0#:

#0 =5x^2-2#

Flip the equation:

#5x^2-2= 0#

Add 2 to both sides:

#5x^2=2#

Divide both sides by 5:

#x^2=2/5#

When we perform the square root operation on both sides, we obtain a negative value and a positive value:

#x=-sqrt(2/5)# and #x=sqrt(2/5)#

Multiply both by 1 in the form of #5/5#:

#x=-sqrt(2/5 5/5)# and #x=sqrt(2/5 5/5)#

#x=-sqrt(10/25)# and #x=sqrt(10/25)#

#x=-sqrt10/5# and #x=sqrt10/5#

Check that both values produce #y = 0#:

#y=5(-sqrt10/5)^2-2# and #y=5(-sqrt10/5)^2-2#

#y=10/5-2# and #y=10/5-2#

#y=2-2# and #y=2-2#

#y=0# and #y=0 larr# this checks

The zeros are #x = -sqrt10/5 and x = sqrt10/5#

Answer 2 · 2018-03-20T23:04:03

To find the answer, set up the problem so #5x^2-2=0#.

To find the answer, set up the problem so #5x^2-2=0#. Then, move everything to one side of the equation, so #x# is by itself.

#5x^2-2=0#

Add 2 to both sides to get:
#5x^2=2#

Then, divide both sides by 5:
#x^2=2/5#

Then, take the square root of both sides:
#sqrt(x^2)=sqrt(2/5)#

The final answer is:
#x=+sqrt(2/5)# and #-sqrt(2/5)#