x2 + 3x - 1 is an expression, NOT an equation so, as it stands, there is nothing to solve.
If the equation was the above expression = 0, then using the quadratic equation,
x = [-3 ± sqrt(9 + 4*1*1)]/2
= [-3 ± sqrt(13)]/2
= -0.30278 and 3.30278 (to 5 dp)
This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========
(2x)ysquared
(x+5)^2 and therefore x = -5
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
x2+5x+6 = (x+2)(x+3) when factored because 2 and 3 are factors of 6
x2+7x+12
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========
(2x)ysquared
-130
(x+5)^2 and therefore x = -5
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
Given the limited information in the question, Z is maximised when x1 or x2 (or both) are maximised. There is no trade-off between x1 and x2 to worry about.
x2+5x+6 = (x+2)(x+3) when factored because 2 and 3 are factors of 6
(2-r)e-rr
You will need to use the FOIL formula to solve this type of equation.
x=4 x=0