4 unknowns needs 4 equations. I'll call the unknown numbers a,b,c,d in that order left to right.
4 + a = b
a + b = c
b + c = d
c + d = 67
let's use substitution to get rid to combine equations and get rid of variables.
If a + b = c then a = c - b
4 + (c - b) = b
4 + c = 2b
If b + c = d then b = d - c
4 + c = 2(d - c)
4 + c = 2d - 2c
4 + 3c = 2d
then we have c + d = 67 so c = 67 - d
4 + 3(67 - d) = 2d
4 + 201 - 3d = 2d
205 = 5d
d = 41
Should be easy now, subtract backwards.
67 - 41 = 26
41 - 26 = 15
26 - 15 = 11
15 - 11 = 4
4, 11, 15, 26, 41, 67
So you need to come up with a perfect square that works for the x coefficients.
like.. (2x + 2)^2
(2x+2)(2x+2) = 4x^2 + 8x + 4
Compare this to the equation given. Our perfect square has +4 instead of +23. The difference is: 23 - 4 = 19
I'm going to assume the given equation equals zero..
So, If we add subtract 19 from both sides of the equation we get the perfect square.
4x^2 + 8x + 23 - 19 = 0 - 19
4x^2 + 8x + 4 = - 19
complete the square and move 19 over..
(2x+2)^2 + 19 = 0
factor the 2 out becomes 2^2 = 4
ANSWER: 4(x+1)^2 + 19 = 0
for a short cut, the standard equation
ax^2 + bx + c = 0 becomes a(x - h)^2 + k = 0
Where "a, b, c" are the same and ..
h = -b/(2a)
k = c - b^2/(4a)
Vertex = (h, k)
this will be a minimum point when "a" is positive upward facing parabola and a maximum point when "a" is negative downward facing parabola.
D. 1/3
A cube root can be written as an exponent: 1/3
![\sqrt[3]{x} = x^{\frac{1}{3} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D)
2x^-3= 2/x^3
you can't multiple 2x to the power of -3 because is not correct. so you need to change the equation for it to give you an answer
