McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 29 Page 218

Practice makes perfect
a
In the Quadratic Formula, b^2-4ac is the discriminant. x=- b±sqrt(b^2-4ac)/2a Having the equation written in standard form, we can now identify the values of a, b, and c. x^2-2x+9=0 ⇔ 1x^2+( -2)x+ 9=0 Let's evaluate the discriminant.
b^2-4ac
( -2)^2-4( 1)( 9)
â–Ľ
Simplify
4-4(1)(9)
4-4(9)
4-36
-32
b If we want to know the number of real solutions we only need to work with the discriminant. Remember that if the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions. b^2-4ac=-32 Since the discriminant is -32, the quadratic equation has no real solutions. However, it does have two imaginary roots.
c We will use the Quadratic Formula to solve the given quadratic equation.
ax^2+ bx+ c=0 ⇔ x=- b± sqrt(b^2-4 a c)/2 a We know that a= 1, b= -2, and c= 9. and that the discriminant b^2-4ac=-32. Let's substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- ( -2)±sqrt(-32)/2( 1)
â–Ľ
Solve for x and Simplify
x=2±sqrt(-32)/2(1)
x=2±sqrt(-32)/2
x=2 ± (sqrt(-1) * sqrt(32))/4
x=2 ± (i * sqrt(32))/4
x=2 ± (sqrt(16) * sqrt(2))i/4
x=2 ± 4 sqrt(2)i/4
x=1 ± 2 sqrt(2)i/2
Using the Quadratic Formula, we found that the two imaginary solutions of the given equation are x= 1 - 2 sqrt(2)i2 and x= 1 + 2 sqrt(2)i2.