(a) what should be add to k^2-10/11k to make it a perfect square? (b)solve the equation v^2-v-1=0?
v - 1/2 = ±√5/2
v = (1±√5)/2
(half of 10/11)^2 = ( 5/11)^2 =25/121
k^2 - 10/11 k + 25/121
v^2 - v = 1 divide -1 by 2 and then square and add both sides
v^2 - 1 v + 1/4 = 1 + 1/4
(v-1/2)(v-1/2) = 1.25
v - .5 = +/-1.25
(a) To make the expression "k^2 - (10/11)k" a perfect square, we need to determine what terms can be added to it.
The main idea is to look for a term that, when added to the expression, creates a perfect square trinomial.
To find this term, we can look at the coefficient of the linear term (k term) in the given expression, which is -10/11.
To make the expression a perfect square trinomial, we need to divide the coefficient of the linear term by 2 and square it. So, we divide -10/11 by 2 and square the result:
(-10/11) / 2 = -5/11 (Dividing by 2)
(-5/11)^2 = 25/121 (Squaring the result)
Therefore, adding 25/121 to the given expression will make it a perfect square. Thus, the expression "k^2 - (10/11)k + 25/121" is a perfect square.
(b) To solve the equation v^2 - v - 1 = 0, we can use the quadratic formula or factorization.
Using the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -1, and c = -1. Plugging these values into the formula, we have:
v = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))
v = (1 ± √(1 + 4))/2
v = (1 ± √5)/2
Therefore, the solutions to the equation v^2 - v - 1 = 0 are v = (1 + √5)/2 and v = (1 - √5)/2.