SquaresIf a+b=8 and a*b=12, what is the sum of squares of a and b? Find also the difference a^2-b^2.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The sum of squares is:

a^2 + b^2

If we'll complete the square, we'll get:

a^2 + b^2 + 2ab - 2ab = (a+b)^2 - 2ab (1)

We'll substitute in (1) the values of the sum a+b and the value of the product a*b.

a^2 + b^2 = 8^2 - 2*12

a^2 + b^2 = 64 - 24

a^2 + b^2 = 40

Now, we'll calculate the difference of squares:

a^2 - b^2

We'll write the difference of squares as a product:

a^2 - b^2 = (a-b)(a+b)

To substitute the factor a-b by it's value, we'll have to calkculate a and b.

We'll have the system of equations:

a + b = 8

a = 8 - b (2)

a*b = 12 (3)

We'll substitute a = 8 - b in (3):

(8 - b)*b = 12

8b - b^2 = 12

We'll subtract 12 both sides:

- b^2 + 8b - 12 = 0

We'll multiply by -1:

b^2 - 8b + 12 = 0

b1 = [8+sqrt(64-48)]/2

b1 = (8+4)/2

b1 = 6

b2 = 2

a1 = 8 - b1

a1 = 8 - 6

a1 = 2

a2 = 6

The system is symmetric.

a - b = 2-6 = -4

a-b = 6-2 = 4

a^2 - b^2 = (a-b)(a+b) = (-4)(8) = -32

or

a^2 - b^2 = 32