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Quadratic equations for Edexcel GCSE Maths

Quadratic equations

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.

Solving quadratics

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.

Completing the square

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.

The quadratic formula

1

Solve xยฒ + 17x โˆ’ 3 = 2x โˆ’ 39.

Rearranging gives xยฒ + 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 = โˆ’12 or x = โˆ’3.

x = {โˆ’12, โˆ’3}

Solve xยฒ + 17x โˆ’ 3 = 2x โˆ’ 39.

2

A rectangle has sides xยฒ and 3 โˆ’ 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 xยฒ + xยฒ + 3 โˆ’ 2x + 3 โˆ’ 2x = 2xยฒ โˆ’ 4x + 6.
Since the perimeter is 6, 2xยฒ โˆ’ 4x + 6 = 6.
Rearranging gives 2xยฒ โˆ’ 4x = 0.
Factorising gives 2x(x โˆ’ 2) = 0.
Equating each factor to 0 gives 2x = 0 or x โˆ’ 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

A rectangle has sides xยฒ and 3 โˆ’ 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.

3

Use the quadratic formula to solve xยฒ + 3x โˆ’ 4 = 0.

x = (โˆ’3 ยฑ โˆš(3ยฒ โˆ’ 4 ร— 1 ร— (โˆ’4)))/(2 ร— 1).
x = (โˆ’3 ยฑ 5)/2

x = {โˆ’4, 1}

Use the quadratic formula to solve xยฒ + 3x โˆ’ 4 = 0.

4

A rectangle has sides 2x + 11 and 4x โˆ’ 1. Given that the area of the rectangle is 20 unitsยฒ, 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 โˆ’ 1) = 8xยฒ โˆ’ 2x + 44x โˆ’ 11 = 8xยฒ + 42x โˆ’ 11
8xยฒ + 42x โˆ’ 11 = 20
8xยฒ + 42x โˆ’ 31 = 0
x = (โˆ’42 ยฑ โˆš(42ยฒ โˆ’ 4(8)(โˆ’31)))/2(8)
x = (โˆ’42 ยฑ 2โˆš689)/16 = (โˆ’21 ยฑ โˆš689)/8
x = (โˆ’21 โˆ’ โˆš689)/8 would make two of the sides of the rectangle negative, which is impossible, therefore it is rejected. Thus, x = (โˆ’21 + โˆš689)/8 = 0.656 (to 3 significant figures).

0.656

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

5

Solve xยฒ + 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 = โˆ’2 or x = โˆ’1.

x = {โˆ’2, โˆ’1}

Solve xยฒ + 3x + 2 = 0.

End of page

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