# StudySquare

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.

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.

For any quadratic equation in the form ax² + bx + c = 0, the discriminant, b² − 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 remainder theorem states that if a polynomial f(x) is divided by (x − a), the remainder is f(a). The factor theorem is a special case of the remainder theorem; when f(a) = 0, (x − 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 − a) is a factor of f(x), substitute a to f(x) and if it gives 0, (x − a) is a factor.

# 1

Fully factorise 15x³ − 154x² − 119x − 22 given that (x − 11) is a factor of the expression.

Since (x − 11) is a factor, factorising by inspection gives (x − 11)(15x² + 11x + 2).

Factorising the quadratic factor gives (x − 11)(3x + 1)(5x + 2).

(x − 11)(3x + 1)(5x + 2)

# 2

Calculate the remainder of x³ + 7x² + 12x − 23 divided by (x − 9) without a long division.

By the remainder theorem, the remainder when the polynomial is divided by (x − 9) is equal to f(9).

f(9) = 9³ + 7(9)² + 12 × 9 − 23 = 1381

1381

# 3

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}

# 4

Given that (x − 2) is a factor of 8x³ + 34x² − 37x − 126, solve 8x³ + 34x² − 37x − 126 = 0.

Since (x − 2) is a factor, factorising by inspection gives y = (x − 2)(8x² + 50x + 63).

Factorising the quadratic factor gives y = (x − 2)(4x + 7)(2x + 9).

x − 2 = 0 or 4x + 7 = 0 or 2x + 9 = 0

x = 2 or x = −7/4 or x = −9/2

x = {−9/2, −7/4, 2}

# 5

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

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