Here's my question: Determine if each system has one, two, or no solutions. (Show your work - 3 marks each)
a)y=2x^2-2x+1 and y=3x-5
b)y=x^2+3x-16 and y=-x^2-8x-18
Here's my work on the question:
a)
x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(--2 plus/minus sign (square root -2^2-4(2)(1)/(2(2)
x=(2 plus/minus sign (square root -4-8)/(4)
x=( 2 plus/minus sign (non-real answer)/(4)
x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(--5 plus/minus sign (square root -5^2-4(3)/2(3)
x=(5 plus/minus sign(square root -25-12)/6
x=(5 plus/minus sign(square root -37)/6
x=(5 plus/minus sign (non-real answer)/6
b)
x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(-3 plus/minus sign (square root 3^2-4(1)(-16)/(2(1))
x=(-3 plus/minus sign (square root 9--64)/2
x=(-3 plus/minus sign (square root 73)/2
x=(-3 plus/minus sign (8.54)/(2)
x=4.27
x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(--8 plus/minus sign(square root -8^2-4(-1)(-18)/(2(-1)
x=(8 plus/minus sign(square root-64-72)/(-2)
x=(8 plus/minus sign(square root -136)/(-2)
x=(8 plus/minus sign(non-real answer)/(-2)
Can you please help me where I went wrong and help correct my errors!
first, do a google on special characters to find the ±√ symbols, which you can then copy and paste. All those words are just too ugly for words! You have
y=2x^2-2x+1 and y=3x-5, so set them equal:
2x^2-2x+1 = 3x-5
2x^2-5x+6 = 0
The discriminant (b^2-4ac) is negative, so there are no solutions.
For the other,
y=x^2+3x-16 and y=-x^2-8x-18
x^2+3x-16 = -x^2-8x-18
2x^2+11x+2 = 0
the discriminant is positive, so it has two distinct solutions.
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D2x^2-2x%2B1+%2C+y%3D+3x-5
http://www.wolframalpha.com/input/?i=plot+y%3Dx^2%2B3x-16+%2C+y%3D-x^2-8x-18
To determine if each system has one, two, or no solutions, we need to find the values of x where the two equations intersect (the points where the values of y are the same for both equations). Let's go through the calculations for each system:
a) The system of equations is:
y = 2x^2 - 2x + 1
y = 3x - 5
To find the solutions, we need to set the two equations equal to each other:
2x^2 - 2x + 1 = 3x - 5
Rearranging the equation to standard form, we get:
2x^2 - 2x - 3x + 1 + 5 = 0
2x^2 - 5x + 6 = 0
Now we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation: a = 2, b = -5, c = 6
x = (-(-5) ± √((-5)^2 - 4(2)(6))) / (2(2))
x = (5 ± √(25 - 48)) / 4
x = (5 ± √(-23)) / 4
The square root of a negative number is imaginary, so this system has no real solutions. Thus, the answer is "no solution."
b) The system of equations is:
y = x^2 + 3x - 16
y = -x^2 - 8x - 18
Setting the two equations equal to each other:
x^2 + 3x - 16 = -x^2 - 8x - 18
Rearranging the equation to standard form:
2x^2 + 11x + 2 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation: a = 2, b = 11, c = 2
x = (-(11) ± √((11)^2 - 4(2)(2))) / (2(2))
x = (-11 ± √(121 - 16)) / 4
x = (-11 ± √105) / 4
The square root of 105 is a real number, so this system has two real solutions. Thus, the answer is "two solutions."
To summarize:
a) The system has no solution.
b) The system has two solutions.