Solve the equation by completing the square. Round to the nearest hundredth if necessary.

Question

Solve the equation by completing the square. Round to the nearest hundredth if necessary. x²+5x-10=0

Answer 1

First, move the 10 to the other side. Because the 10 is subtracted, we add.

x² + 5x = 10

Now, take half of the second term's coefficient, square it, and add it to both sides.

(5/2)² = 25/4

x² + 5x + 25/4 = 10 + 25/4

Now, the left side can be factored into a binomial, squared, using that 5/2. I'm going to ignore the right side for the moment.

x² + 5x + 25/4 = (x + 5/2)²

Now, 10 + 25/4 = 40/4 + 25/4 = 65/4, right? Let's put the two sides back together.

(x + 5/2)² = 65/4

x + 5/2 = √65/4

x = √65/4 - 5/2

x = 1.53

Answer 2

• Original: x²+5x-10=0

• Begin by adding 10 to both sides: x²+5x=10

• To complete the square, take the

b value of the equation corresponding

to the general quadratic equation

ax²+bx+c, in this case b=5, divide by

two and square: (b/2)²=(5/2)²=25/4

• Add this value to both sides: x²+5x+(25/4)=10+(25/4)

• Simplify right side and factor left: (x+5/2)(x+5/2)=(65/4)

• Solve for x: (x+5/2)²=(65/4)

x+(5/2)=±√(65)/2

x=(±√(65)-5)/2