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OCR A-level Maths Quadratic equations

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 axΒ² + bx + c = 0, the discriminant, bΒ² βˆ’ 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 βˆ’ a), the remainder is f(a). The factor theorem is a special case of the remainder theorem; when f(a) = 0, (x βˆ’ 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 βˆ’ a) is a factor of f(x), substitute a to f(x) and if it gives 0, (x βˆ’ a) is a factor.

The factor theorem

1

Solve xΒ² + 10x βˆ’ 24 = 0 for x by completing the square.

Solve xΒ² + 10x βˆ’ 24 = 0 for x by completing the square.

2

Given that the quadratic equation xΒ² βˆ’ 4px βˆ’ 2.5 + 7p = 0 has one real root, find the two possible values of p.

Given that the quadratic equation xΒ² βˆ’ 4px βˆ’ 2.5 + 7p = 0 has one real root, find the two possible values of p.

3

The area of some trapezium has an expression of 10xΒ² + 4x. Given that the area of this trapezium is 6 unitsΒ², find the positive value for x using the quadratic formula.

The area of some trapezium has an expression of 10xΒ² + 4x. Given that the area of this trapezium is 6 unitsΒ², find the positive value for x using the quadratic formula.

4

Calculate the remainder of xΒ³ + 7xΒ² + 12x βˆ’ 23 divided by (x βˆ’ 9) without a long division.

Calculate the remainder of xΒ³ + 7xΒ² + 12x βˆ’ 23 divided by (x βˆ’ 9) without a long division.

5

Use the quadratic formula to obtain x for 4(xΒ² + 6) βˆ’ (x βˆ’ 1) = 24(xΒ² + 1).

Use the quadratic formula to obtain x for 4(xΒ² + 6) βˆ’ (x βˆ’ 1) = 24(xΒ² + 1).

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