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How do you solve #2x^2+5=41#?

Answer 1

Savannah Anderson

#x=+-3sqrt2#

#"isolate "2x^2" by subtracting 5 from both sides"#

#rArr2x^2=41-5=36#

#"divide both sides by 2"#

#rArrx^2=36/2=18#

#color(blue)"take the square root of both sides"#

#sqrt(x^2)=+-sqrt18larrcolor(blue)"note plus or minus"#

#rArrx=+-sqrt18=+-3sqrt2#

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Answer 2

Savannah Anderson

To solve the equation 2x^2 + 5 = 41:

Subtract 5 from both sides of the equation: 2x^2 = 36
Divide both sides by 2: x^2 = 18
Take the square root of both sides: x = ±√18
Simplify the square root of 18: x = ±3√2 or x ≈ ±4.24

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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.