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Quadratic equations for OCR A-level Maths

Quadratic equations

This page covers the following topics:

1. Solving quadratics
2. Completing the square
3. The quadratic formula
4. The discriminant
5. The factor theorem

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

For any quadratic equation in the form axยฒ + bx + c = 0, the discriminant, bยฒ โˆ’ 4ac, can be used to determine the number of roots the equation has. When the discriminant is positive, the quadratic equation has two distinct real roots. When the discriminant is 0, the quadratic equation has one real root (or two same roots). When the discriminant is negative, the quadratic equation has no real roots.

Information about the number of real solutions can also be read from graphs. The number common points between a parabola that represents a quadratic function and the x axis is the same as the number of real solutions.

The discriminant

The remainder theorem states that if a polynomial f(x) is divided by (x โˆ’ a), the remainder is f(a). The factor theorem is a special case of the remainder theorem; when f(a) = 0, (x โˆ’ a) is a factor of the polynomial and a is a solution to the polynomial, and vice versa. Thus, if you want to check that (x โˆ’ a) is a factor of f(x), substitute a to f(x) and if it gives 0, (x โˆ’ a) is a factor.

The factor theorem

1

Fully factorise 15xยณ โˆ’ 154xยฒ โˆ’ 119x โˆ’ 22 given that (x โˆ’ 11) is a factor of the expression.

Since (x โˆ’ 11) is a factor, factorising by inspection gives (x โˆ’ 11)(15xยฒ + 11x + 2).
Factorising the quadratic factor gives (x โˆ’ 11)(3x + 1)(5x + 2).

(x โˆ’ 11)(3x + 1)(5x + 2)

Fully factorise 15xยณ โˆ’ 154xยฒ โˆ’ 119x โˆ’ 22 given that (x โˆ’ 11) is a factor of the expression.

2

Calculate the remainder of xยณ + 7xยฒ + 12x โˆ’ 23 divided by (x โˆ’ 9) without a long division.

By the remainder theorem, the remainder when the polynomial is divided by (x โˆ’ 9) is equal to f(9).
f(9) = 9ยณ + 7(9)ยฒ + 12 ร— 9 โˆ’ 23 = 1381

1381

Calculate the remainder of xยณ + 7xยฒ + 12x โˆ’ 23 divided by (x โˆ’ 9) without a long division.

3

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.

4

Given that (x โˆ’ 2) is a factor of 8xยณ + 34xยฒ โˆ’ 37x โˆ’ 126, solve 8xยณ + 34xยฒ โˆ’ 37x โˆ’ 126 = 0.

Since (x โˆ’ 2) is a factor, factorising by inspection gives y = (x โˆ’ 2)(8xยฒ + 50x + 63).
Factorising the quadratic factor gives y = (x โˆ’ 2)(4x + 7)(2x + 9).
x โˆ’ 2 = 0 or 4x + 7 = 0 or 2x + 9 = 0
x = 2 or x = โˆ’7/4 or x = โˆ’9/2

x = {โˆ’9/2, โˆ’7/4, 2}

Given that (x โˆ’ 2) is a factor of 8xยณ + 34xยฒ โˆ’ 37x โˆ’ 126, solve 8xยณ + 34xยฒ โˆ’ 37x โˆ’ 126 = 0.

5

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.

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