7.1 Completing the Square
pg 348 #1
Solve
(3x+1)^2=8
(3x+1)=+-root*(8)
3x=-1+-root()(8)
x=(-1+-root()(8))/3
pg 348 #5a
Without solving the equation, determine the nature of its roots.
5x^2+7x+2=0
Discriminant =
b^2-4ac
49-4(5)(2)=49-40=9
The discriminant>0 so there are two real roots that are irrational
pg 324 #1a
Solve
(x+3)^2-5(x+3)+4=0
z=x+3
z^2=(x+3)^2
z^2-5z+4=0
(z-4)(z-1)=0
z=4, z=1
x+3=4, x+3=1
x=1, x=-2
pg348 #8
Graph the parabola. Label the vertex an axis of symmetry
y+3=-1/2(x-2)^2
vertex = (2,-3)
Axis of symmetry is x=2
a<1, so it will be concave down.
pg 348 #14
Find the dimensions of a rectangle of greatest area whose perimeter is 20cm
2l+2w = 20
l+w=10
l=10-w
lw=A
w(10-w)=A
10w-w^2=A
max is at (h,k) h =-b/2a
(-10)/(2(-1))=5=h
l=5 and w = 5
pg310 #13
Solve by completing the square
x^2-2x-5=0
x^2-2x=5
x^2-2x+1=5+1
(x-1)^2=6
x-1=+-root()(6)
x=1+-root()(6)
pg 348 #5b
Without solving the equation, determine the nature of its roots.
3x^2-4x+2=0
Discriminant =
b^2-4ac
16-4(3)(2)=16-24=-8
The discriminant<0 so there are two imaginary roots
pg 348 #7a
Solve each equation with quadratic form over complex numbers.
x^4+x^2-12=0
f(x)=z=x^2
(f(x))^2=z^2=x^4
z^2+z-12=0
(z+4)(z-3)=0
z=3,z=-4
x^2=3,x^2=-4
x=+-root()(3),x=+-root()(-4)=+-2i
pg 348 #9
Find an equation in the form
y-k=a(x-h)^2
with vertex (-2,5) and containing point (2,9)
vertex is (-2,5) so h=-2 and k=5
y-5=a(x+2)^2
plug in (2,9) for x and y
9-5=a(2+2)^2
4=a(4)^2
4=16a
a=1/4
y-5=1/4(x+2)^2
pg 344 #12
A ball is thrown vertically upward with an initial speed of 80ft/s. Its height after t seconds is given by
h=80t-16t^2
a) How high does the ball go?
b)When does the ball hit the ground?
a)To find the maximum height we can find the vertex (h,k).
h=-b/(2a)=-80/(2*-16)=80/32=5/2
k=80(5/2)-16(5/2)^2
k=200-16(25/4)=200-100=100ft
b)to find when the ball hits the ground we need to find the zeros for the function.
0=80t-16t^2=t(80-t)
t=0,80
The ball starts at 0ft high at 0 and then hits the ground after being thrown at 80sec
pg 310 #17
Solve by completing the square
p^2+20p+200=0
p^2+20p=-200
p^2+20p+100=-200+100
(p+10)^2=-100
p+10=+-root(-100)
p=-10+-10i
pg 348 #3
Use the quadratic formula to solve
4x^2-3x+2=0
x=(3+-root()(9-4(4)(2)))/(2(4))
x=(3+-root()(9-32))/8
x=(3+-root()(-23))/8
pg 324 #13
Solve
((1+x)/2)^2-3((1+x)/2)=18
z=((1+x)/2)
z^2=((1+x)/2)
z^2-3z-18=0
(z-6)(z+3)=0
z=6,z=-3
((1+x)/2)=6,((1+x)/2)=-3
1+x=12,1+x=-6
x=11,x=-7
pg.348 #10
Graph the function
f(x)=2x^2-4x+1
After putting it into the form
y-k=a(x-h)^2
(Hint: Complete the square)
Complete the square
y=2x^2-4x+1
y-1=2(x^2-2x)
y-1+2(1)=2(x^2-2x+1)
y+1=2(x-1)^2
Vertex is (1,-1) and a > 0 so it will be concave up
pg 348 #12
Find a quadratic equation with integral coefficients having roots
(1+-root()(3))/4
sum of roots = -b/a
product of roots = c/a
(1+root()(3))/4+(1-root()(3))/4=2/4=4/8=c/a
-(1+root()(3))/4*(1-root()(3))/4=-(1-3)/16=2/16=1/8=-b/a
c=4,b=-1,a=8
8x^2-x+4=f(x)
pg348 #2
Solve by completing the square
2x^2+6x+3=0
x^2+3x+3/2=0
x^2+3x=-3/2
x^2+3x+9/4=9/4-3/2
(x+3/2)^2=(9-6)/4=3/4
x+3/2=+-root()(3/4)=+-root()(3)/2
x=(-3+-root()(3))/2
pg 348 #4
Two positive real numbers have a sum of 7 and a product of 11. Find the numbers.
x+y=7
xy=11
y=7-x
x(7-x)=11
7x-x^2-11=0
-x^2+7x-11=0
x=(-7+-root()(49-4(-1)(-11)))/(2*-1)
x=(-7+-root()(49-44))/(-2)
x=(-7+-root()(5))/(-2)
pg 348 #7b
Solve each equation with quadratic form over complex numbers.
x^(-2)-2x^(-1)-1=0
f(x)=z=x^(-1)
(f(x))^2=z^2=x^(-2)
z^2-2z-1=0
z=(2+-root()(4-4(1)(-1)))/2=(2+-root()(8))/2=1+-root()(2)
x^-1=1/x=1+-root()(2)
x=1/(1+-root()(2))
x=-1+-root()(2)
pg 336 #34
Find the vertex, the domain, the range, and zeros of the function
f(x)=2(x-5)^2-8
can be rewritten as
y+8=2(x-5)^2
Vertex is then (5,-8)
The Domain is all real numbers
the function is concave up because a>0 so -8 is the minimum and the range is >=-8
The zeros are
0+8=2(x-5)^2
4=(x-5)^2
+-2=x-5
5+-2=x=7,3
pg 348 #13
Find a quadratic function
f(x)=ax^2+bx+c
having a minimum value -9 and zeros 1/2 and -5/2
sum of roots = -b/a
product of roots = c/a
(1/2)+(-5/2)=-4/2=-2=c/a
(1/2)(-5/2)=(-5/4)=-b/a
a(x^2+(5x)/4-2)=y
minimum at (-1,-9)
a((-1)^2+(-5)/4-2)=-9
a((4-5-8)/4)=-9
a((-9/4))=-9
a=4
4(x^2+(5x)/4-2)=y
4x^2+5x-8=y
pg 310 #33
Solve by completing the square
2x(x-4)=3(1-x)
2x^2-8x=3-3x
2x^2-5x=3
x^2-(5x)/2=3/2
x^2-(5x)/2+(25/16)=3/2+25/16
(x-5/4)^2=24/16+25/16=49/16
x-5/4=+-root()(49/16)=+-7/4
x=(5+-7)/4
x=3,-1/2
pg. 348 #6
Find all real values of k for which
2x^2+kx+3=0
Has a double root
Discriminant must =0 to have a double root
D=0=b^2-4ac
0=k^2-4(2)(3)=k^2-24
k=+-root()(24)=+-2root()(6)
pg324 #15
Solve
2/y+1/root()(y)=1
z=1/root()(y)
z^2=1/y
2z^2+z-1=0
(2z-1)(z+1)=0
z=-1,z=1/2
1/root()(y)=-1,1/root()(y)=1/2
root()(y)=-1, root()(y)=2
y=4
pg348 #11
Find the domain, range, and zeros of
g(x)=x^2-6x+4
domain is all real numbers
vertex is
-b/(2a)=6/2=3=h
(3)^2-6(3)+4=9-18+4=-5=k
Minimum is -5
Range is >=5
zeros
0=x^2-6x+4
x=(6+-root()(36-4(1)(4)))/2
x=(6+-root()(20))/2
x=3+-root()(5)
pg 347 #15
Find a quadratic function for the parabola having minimum value -2 and x-intercepts 1 and -3.
minimum value is -2 and h is halfway between 1 and -3. So h = -1 and vertex is (-1,-2)
with that you can input into the vertex form
y+2=a(x+1)^2
To find a simply plug in (1,0) or (-3,0) into the equation.
0+2=a(1+1)^2
2=4a
a=1/2
y+2=1/2(x+1)^2
From there we can multiply it out to get the basic form
ax^2+bx+c=y
y+2=1/2(x^2+2x+1)
y=1/2x^2+x+1/2-4/2
y=1/2x^2+x-3/2