Quadratic expressions

\(9x^2-4=(3x)^2-2^2\), so it is a difference of squares and factors as \((3x-2)(3x+2)\). A and D are perfect-square trinomials. Exam tip: first check whether both terms are squares.

Expanding \((px+q)^2\) gives \(p^2x^2+2pqx+q^2\). Thus \(a=p^2\), \(b=2pq\), and \(c=q^2\), so \(b^2=4ac\). Exam tip: square the middle coefficient and compare it with \(4ac\).

The product of identical binomials, such as \((x+3)(x+3)=(x+3)^2\), is a perfect-square trinomial. A difference of squares has factors \((a+b)(a-b)\), which are not identical. Exam tip: identical binomial factors signal a perfect square.

In \(4x^2-3x+6\), the highest power of \(x\) is 2, and every exponent of \(x\) is a non-negative integer. Hence, it is a quadratic polynomial in \(x\). \(x^3-2x+1\) has degree 3, so it is cubic. \(5x+\frac{2}{x}\) contains \(x^{-1}\), while \(\sqrt{x}+x\) contains \(x^{1/2}\); therefore, neither is a polynomial. Exam tip: in a polynomial, variable exponents must be non-negative integers such as 0, 1, 2, or 3.

(-4-5=-9) and ((-4)(-5)=20). In exams, when constant is positive and middle term is negative, use two negative signs.

A quadratic \(ax^2+bx+c\) has equal zeroes when its discriminant \(b^2-4ac\) is zero. For option A, \(36-4\times1\times9=0\). Option B has discriminant 4, so its zeroes are distinct. Exam tip: check whether the discriminant is zero for equal zeroes.

Here \(a=1, b=10, c=25\), and \(b^2=100=4ac\); hence \(x^2+10x+25=(x+5)^2\), a perfect square. Exam tip: halve the middle-term coefficient and square it.

In option A, \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\). Its middle term is \(-24pq=-2(3p)(4q)\). Hence, \(9p^2-24pq+16q^2=(3p-4q)^2\), so it is a perfect square. In option B, the middle term is \(-20pq\), not the required \(-24pq\). Exam tip: take the square roots of the first and last terms, then check whether the middle term is \(\pm2ab\).

Use \((a+b)^2=a^2+2ab+b^2\). With \(a=x\) and \(b=4\), the middle term is \(2\times x\times4=8x\), so the result is \(x^2+8x+16\). Adding only the squares misses the cross term. Exam tip: always check the middle term.

In \(x^2+\frac{3}{x}+2\), the term \(\frac{3}{x}=3x^{-1}\) has a negative power of \(x\). A polynomial cannot contain negative exponents, so it is not quadratic. An irrational coefficient such as \(\sqrt{2}\) is allowed. Exam tip: check powers first.

For \(k=-3\), the coefficient of \(x^2\) becomes \(k^2-9=0\), while the coefficient of \(x\) is \(-6\). Hence the expression is linear. Exam tip: for a linear expression, the \(x^2\) coefficient must be zero but the \(x\) coefficient must be non-zero.

Here, \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\). The middle term is \(-24ab=-2(3a)(4b)\), so the expression equals \((3a-4b)^2\). In option B, the last term should be \(16b^2\). Exam tip: check the square roots of the first and last terms first.

The other factor is (x+4) because (6\cdot4=24), so (p=6+4=10). In exams, find the other factor from the constant term.

\(9x^2-24x+16=(3x-4)^2\) because the middle term is \(2\times3x\times(-4)=-24x\). In option B, the constant should be 16, not 15. Exam tip: use square roots of the first and last terms to verify the middle term.

If \(x^2+px+q=(x+a)^2\), expanding gives \(x^2+2ax+a^2\). Thus \(p=2a\) and \(q=a^2\), so \(p^2=4q\). Exam tip: compare the square of the middle coefficient with four times the constant term.

\(9a^2-24ab+16b^2=(3a-4b)^2\), using \((x-y)^2=x^2-2xy+y^2\) with \(x=3a\) and \(y=4b\). The middle term is \(-2\times3a\times4b=-24ab\). In option D, the first and last terms are \((3a)^2\) and \((4b)^2\), but its middle term is \(-16ab\), so it is not a perfect square. Exam tip: take square roots of the first and last terms and check whether the middle term equals \(\pm2xy\).

Half of (-14) is (-7), and ((-7)^2=49). In exams, even if the sign is negative, the square is positive.

(9x^2=(3x)^2) and (16=4^2), so the middle coefficient is (2\cdot3\cdot4=24). In exams, view the negative sign and coefficient separately.

\(x^2-y^2\) is a difference of squares: \((x+y)(x-y)=x^2-y^2\). On multiplication, \(+xy\) and \(-xy\) cancel. In exams, look for a minus sign between two square terms.

A quadratic expression must have highest power 2, so the coefficient of \(x^2\) must satisfy \(a\ne0\). The values of b or c may be zero. Exam tip: first check that the leading coefficient is non-zero.