Algebra Assignment

Part 1

Completing the Square

If a quadratic function is one of the form (x ± a)2 = c, then we can solve it by taking the square root of each side. In an equation of this form, the left hand side is a perfect square: the square of a linear expression in x. so if a quadratic equation does not factor readily, and then we can solve it by completing square (Stewart, Redlin & Watson, 2011). To make x2 + bx a perfect square, add (b/2)2, the square of the half of the coefficient of x. this gives the perfect square.

Solution of x2 + 6x - 8 = 0

x2 + 6x - 8 = (x2 + 6x) - 8

Half of 6 is 3 and its square is 9 therefore, adding and subtracting 9 on right hand side of the equation

x2 + 6x - 8 = (x2 + 6x + 9 - 9) - 8 = (x2 + 6x + 9) - 9 - 8

= (x2 + 6x + 9) - 17 … (1)

Where

x2 + 6x + 9 = (x)2 + 2(3)(x)2+ (3)2

= (x + 3)2

Substituting this in equation 1

x2 + 6x - 8 = (x + 3)2 - 17

or

(x + 3)2 - 17 = 0

(x + 3)2 = 17

Taking square root on both sides

v(x + 3)2 = ±v17

As square root is power ½; therefore ½ × 2 becomes 1

x + 3 = ±v17

x = -3 ±v17

Either x = -3 +v17

x = 1.123

Or x = -3 - v17

x = -7.123

Verification

Substituting x = 1.123 in given equation

(1.123)2 + 6(1.123) - 8 = 0.00

Substituting x = -7.123 in given equation

(-7.123)2 + 6(-7.123) - 8 = 0.00

Part 2

Solving 2y2 - 3y - 6 = 0 by quadratic formula

Quadratic Formula: given an equation ax2 + bx - c = 0, the descriminant ...