1. What are the solutions of 0=x squared + 4x - 5 ?
A 4 & 5
B -5 & 1
C.-4 & 5
D. 5 & -1
2.What are the solutions of 0=9x squared - 36 ?
A -6 & 6
B. -4 & 4
C. -3 & 3
D -2 & 2
3. A rectangle with an area of 124 cm squared has a length that is 4 times the width. How long is the width? (round your answer to the nearest 10th)
A 5.6 cm
B. 11.1 cm
C. 22.3 cm
D. 44.5 cm
A 4 & 5
B -5 & 1
C.-4 & 5
D. 5 & -1
2.What are the solutions of 0=9x squared - 36 ?
A -6 & 6
B. -4 & 4
C. -3 & 3
D -2 & 2
3. A rectangle with an area of 124 cm squared has a length that is 4 times the width. How long is the width? (round your answer to the nearest 10th)
A 5.6 cm
B. 11.1 cm
C. 22.3 cm
D. 44.5 cm
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1) If you factor it out you get (x-1)(x+5)
Solving for x using the zero product principle that one of the two terms has to be zero, either x=-5 or x=1
(B)
2) D
That one is easier than the first one. Just add 36 to the left hand side and right hand side. so 9x^2 = 36 then divide by 9 on both sides, and take the square root of both sides. Taking the square root always yields a plus or minus value.
3) So Length=4(width)
and we know that length x width =the area of the rectangle.
so plug in 4W for L in the equation of L*W=area
this gives us 4W*W=area or 4W^2=124
if you solve for W you get (A.)
Solving for x using the zero product principle that one of the two terms has to be zero, either x=-5 or x=1
(B)
2) D
That one is easier than the first one. Just add 36 to the left hand side and right hand side. so 9x^2 = 36 then divide by 9 on both sides, and take the square root of both sides. Taking the square root always yields a plus or minus value.
3) So Length=4(width)
and we know that length x width =the area of the rectangle.
so plug in 4W for L in the equation of L*W=area
this gives us 4W*W=area or 4W^2=124
if you solve for W you get (A.)
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1) x^2 + 4x -5
1 5
1 -1
(1)(-1) = -1
(1)(5) = 5
5 + (-1) = 4 the number that is in the middle
solution:
(1x +5)(1x-1)
set each equal to zero
1x +5 = 0
1x = -5
1x-1 = 0
1x = 1
answer: -5 and 1
same for second question
9x^2 -36
3 6
3 -6
(3)(-6) = -18
(3)(6) = 18
(-18)+18 = 0 the number that you do not see which is 0x
(3x +6)(3x-6)
set each equal to zero
3x+6 = 0
3x=-6
divide by three
x = (-6)/(3) = -2
3x-6 = 0
3x = 6
divide by 3
x= (6)/3 = 2
answer: -2 and 2
for the third question you know that area = length times width
area= 124cm^2
but length = 4(width)
plug that into the area equation and then it becomes:
area = (4W)(W)
A = 4W^2
124 = 4W^2
divide by four
124/4 = W^2
take the square root of each side
square root(124/4) = W^2
your answer would be the positive square root. every time you take a square root you get a positive and negative answer, but in this case you only take the positive square root because the width had to be a positive number. Have you hear of a negative length? (it's physically impossible)
1 5
1 -1
(1)(-1) = -1
(1)(5) = 5
5 + (-1) = 4 the number that is in the middle
solution:
(1x +5)(1x-1)
set each equal to zero
1x +5 = 0
1x = -5
1x-1 = 0
1x = 1
answer: -5 and 1
same for second question
9x^2 -36
3 6
3 -6
(3)(-6) = -18
(3)(6) = 18
(-18)+18 = 0 the number that you do not see which is 0x
(3x +6)(3x-6)
set each equal to zero
3x+6 = 0
3x=-6
divide by three
x = (-6)/(3) = -2
3x-6 = 0
3x = 6
divide by 3
x= (6)/3 = 2
answer: -2 and 2
for the third question you know that area = length times width
area= 124cm^2
but length = 4(width)
plug that into the area equation and then it becomes:
area = (4W)(W)
A = 4W^2
124 = 4W^2
divide by four
124/4 = W^2
take the square root of each side
square root(124/4) = W^2
your answer would be the positive square root. every time you take a square root you get a positive and negative answer, but in this case you only take the positive square root because the width had to be a positive number. Have you hear of a negative length? (it's physically impossible)
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1. B. (x+5) (x-1) so... x= -5 x= 1
2. D. (3x+6) (3x-6) so... x= -2 x=2
3. A. 4x^2= 124 -> x^2=31 -> x=5.6
2. D. (3x+6) (3x-6) so... x= -2 x=2
3. A. 4x^2= 124 -> x^2=31 -> x=5.6