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How do you solve #w^2 = 4# where w is a real number?

Answer 1

To solve this, you have to think about what get rids of a square root.

Think about the number 16...#16=4^2#. But #sqrt(16)=4#. Can you see how the square root gets rid of the square exponent?

Another way to think about it this is that taking a sqare root of something means the same thing as taking that number to the power of #1/2#. So, #sqrt(16)=sqrt(4^2)=(4^2)^(1/2)# and when you have an exponent to an exponent you multiply the exponents so #(4^2)^(1/2)=4^1=4#

#w^2=4# Square root both sides to get x alone: #sqrt(w^2)=sqrt(4)# #w=+-2#

Notice that w can equal two real numbers: +2 and -2. The positive root is quite intuitive- But the negative root also holds true, because when you mutiply two negative numbers, you get a positive number. #(-2)^2=(-2)*(-2)=4#

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Answer 2

Jameson Allen

To solve the equation w^2 = 4 where w is a real number, you can take the square root of both sides of the equation. This gives two solutions: w = 2 and w = -2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.