Simplify first, then decide

The idea

An equation is simplified before deciding whether it is quadratic: one that looks quadratic may not reduce to ax² + bx + c = 0, and one that looks cubic may reduce to it.

Does it have an x²?

Look at x(x + 1) + 8 = (x + 2)(x − 2). The left side has x × x, and the right side has x × x too.

Is this a quadratic equation?

• Yes: it has x² in it
• No: it is not quadratic
• We cannot tell until we simplify it

Look, simplify, decide

For each equation: take a first look, then press to simplify it line by line, then decide whether it is quadratic. The card tells you if the first look was fooled.

Simplify before you decide

An equation is quadratic only if, after you simplify it, it can be written as ax² + bx + c = 0 with a ≠ 0. Always simplify first.

In (ii), the x² on the left and the x² on the right cancelled, and the equation became x + 12 = 0. It *looked* quadratic and is not.

In (iv), the x³ on both sides cancelled, and 6x² + 12x + 12 = 0 was left. It *looked* cubic and is quadratic.

Notes

Simplify first. The equation is quadratic only if it reduces to ax² + bx + c = 0 with a ≠ 0. A quadratic-looking equation may lose its x², and a cubic-looking one may lose its x³.

Check yourself

Is x² + 3x = x² − 5 a quadratic equation?

Is (x + 1)³ = x³ − 3x + 4 a quadratic equation? (Hint: (x + 1)³ = x³ + 3x² + 3x + 1.)

Carry every term of x(x + 4) = 2x² − 5 to the left side. What is a, the number in front of x²?

Answer: -1

x² + 4x − 2x² + 5 = 0 gives −x² + 4x + 5 = 0. So a = −1, which is not 0: the equation is quadratic.

Why do we simplify before deciding whether an equation is quadratic?

• No: the x² terms cancel, leaving 3x + 5 = 0 — correct. Yes! Carry x² across: 3x = −5. There is no x² left.
• Yes: it has x² in it. Look again after simplifying. The x² on each side cancels.
• Yes: it has x² on both sides. Two x² terms that cancel leave no x² at all.
• Yes: it simplifies to x² + 2x − 1 = 0 — correct. Yes! x³ + 3x² + 3x + 1 = x³ − 3x + 4 gives 3x² + 6x − 3 = 0, and dividing by 3 gives x² + 2x − 1 = 0.
• No: it has x³ in it. The x³ on both sides cancels. Simplify and look at what is left.
• No: it simplifies to a line. After x³ cancels, 3x² + 6x − 3 = 0 is left. It still has x².
• The first look can be fooled: x² or x³ terms may cancel — correct. Yes! (ii) lost its x², and (iv) lost its x³.
• The first look is always right, simplifying only makes it neater. The first look was fooled in (ii) and (iv).
• The highest power on the first look is always the degree. Not when terms cancel. (iv) shows x³, but its degree is 2.