Quadratic expressions

A quadratic expression must retain an \(x^2\) term, so its coefficient must satisfy \(a\ne0\). The coefficients \(b\) and \(c\) may be zero. Exam tip: check the coefficient of the highest power first.

In \(4x^2-7x\), the highest power of x is 2, so it is quadratic, and its constant term is 0. \(4x^2-7\) has constant term −7. Exam tip: the constant term contains no x.

The numbers (-2) and (-3) have sum (-5) and product (6). Exam tip: match signs with the constant term.

A quadratic expression has the form \(ax^2+bx+c\), where \(a\ne0\). Its graph is a parabola, opening upward or downward according to the sign of \(a\). A straight line represents a linear expression. Exam tip: an \(x^2\) term indicates a parabola.

(x^2+4x+4=(x+2)^2) and (x^2-4=(x-2)(x+2)). Exam tip: distinguish similar-looking quadratics by identity.

Here (-2-3=-5) and ((-2)(-3)=6). Exam tip: for a positive constant and negative middle term, take both numbers negative.

Expanding ((2x-1)(x-3)) gives (2x^2-6x-x+3). Exam tip: verify the middle term by expansion.

Since (36=6^2), the middle term will be (2\cdot6\cdot x=12x). Exam tip: double the square root of the constant term.

Here (4x^2=(2x)^2) and (9=3^2), so it is a difference of two squares. Exam tip: apply the (a^2-b^2) identity quickly.

It is in the form (x^2-2\cdot x\cdot1+1^2). Exam tip: in ((a-b)^2), the middle term is negative.

A quadratic expression must have highest power 2, so the coefficient \(a\) of \(x^2\) cannot be zero. If \(a=0\), the expression becomes at most linear. Exam tip: first check the coefficient of the highest-power term.

The term without a variable is the constant term. In exams write the constant term with its sign.

(9=3^2) and (6x=2\cdot x\cdot3). In exams take the square root of the last term and match the middle term.

The middle term (-10a=-2\cdot a\cdot5), so the sign is negative. In exams the sign of the middle term is decisive.

(3+5=8) and (3\cdot5=15). In exams match the sum with the middle coefficient and the product with the constant term.

(-3+(-4)=-7) and ((-3)(-4)=12). In exams use two negative signs when the constant is positive and the middle term is negative.

(5+(-4)=1) and (5\cdot(-4)=-20). In exams look for opposite signs when the constant is negative.

In ((3x+4)(x+2)), the cross terms are (6x) and (4x). In exams their sum should be the middle term.

Here, \(4x^2=(2x)^2\) and \(9=3^2\). Therefore, \(4x^2-9=(2x)^2-3^2\) is a difference of squares. Applying \(a^2-b^2=(a-b)(a+b)\) gives \((2x-3)(2x+3)\). Option \((2x-3)^2\) expands with a middle term \(-12x\), so it is not correct. Exam tip: for a difference of squares, take the square roots of both terms and write their difference and sum as factors.

(9y^2=(3y)^2), (16=4^2), and the middle term is (-24y). In exams match both square terms and the middle term.