Question

Task 1

Answer 1

Answer:

part 1 : y = x^2 + 3x + 5

y = x + 13

solution:

x^2 + 3x + 5 = x + 13

x^2 + 2x -8 = 0

(x+4)(x-2) = 0

x = -4 or x = 2

if x = -4, y = 9

if x = 2 , y = 15

part 2 : x^2 + 2x + 1 = 0

(x + 1)^2 = 0

x = -1

y = 1

task 2

part 1 : Let our two digit number be 26

Next, we rewrite it as a difference of two numbers

26=32-6

We want to square 26 using the identity:  

In this case,

x=32

y=6

Substituting into the identity

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