# StudySquare

This page covers the following topics:

1. Solving quadratics

2. Completing the square

3. The quadratic formula

Quadratics can be solved by equating one side of the equation to zero and factorising. For a product to be equal to zero, one or both of the factors must be equal to zero. Therefore, the quadratic equation can be solved by equating each factor to zero and solving for the variable.

Quadratic equations that cannot be factorised can be solved by completing the square. Completing the square results in a square plus a term, which can be solved by moving the additional term to the other side of the equation and square rooting both sides.

A common method to solve quadratics is by using the quadratic formula. Substituting coefficients a, b, c of the quadratic equation into the formula gives its solutions.

# 1

Solve x² + 17x − 3 = 2x − 39.

Rearranging gives x² + 15x + 36 = 0.

Factorising gives (x + 12)(x + 3) = 0.

Equating each factor to 0 gives x + 12 = 0 or x + 3 = 0.

Thus, x = −12 or x = −3.

x = {−12, −3}

# 2

A rectangle has sides x² and 3 − 2x. Given that the perimeter of the rectangle is 6, find x. The perimeter of a shape is a sum of the lengths of all of its sides.

The perimeter of the rectangle is given by x² + x² + 3 − 2x + 3 − 2x = 2x² − 4x + 6.

Since the perimeter is 6, 2x² − 4x + 6 = 6.

Rearranging gives 2x² − 4x = 0.

Factorising gives 2x(x − 2) = 0.

Equating each factor to 0 gives 2x = 0 or x − 2 = 0.

Thus, x = 0 or x = 2.

When x = 0, two sides of the rectangle would be 0, which is impossible.

When x = 2, one of the sides of the rectangle is less than 0, which is impossible.

Therefore, there are no solutions.

no solutions

# 3

Use the quadratic formula to solve x² + 3x − 4 = 0.

x = (−3 ± √(3² − 4 × 1 × (−4)))/(2 × 1).

x = (−3 ± 5)/2

x = {−4, 1}

# 4

A rectangle has sides 2x + 11 and 4x − 1. Given that the area of the rectangle is 20 units², calculate x using the quadratic formula to 3 significant figures. The area of a rectangle is the multiplication product of its two perpendicular sides.

area of rectangle = (2x + 11)(4x − 1) = 8x² − 2x + 44x − 11 = 8x² + 42x − 11

8x² + 42x − 11 = 20

8x² + 42x − 31 = 0

x = (−42 ± √(42² − 4(8)(−31)))/2(8)

x = (−42 ± 2√689)/16 = (−21 ± √689)/8

x = (−21 − √689)/8 would make two of the sides of the rectangle negative, which is impossible, therefore it is rejected. Thus, x = (−21 + √689)/8 = 0.656 (to 3 significant figures).

0.656

# 5

Solve x² + 3x + 2 = 0.

Factorising gives (x + 2)(x + 1) = 0.

Equating each factor to 0 gives x + 2 = 0 or x + 1 = 0.

Thus, x = −2 or x = −1.

x = {−2, −1}

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