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 \(\pm2xy\).

The direct answer is A: \(c=49\). A monic quadratic \(x^2+bx+c\) is a perfect square when it has the form \((x+b/2)^2\), so its constant term is \((b/2)^2\). Here the middle coefficient is \(-14\). Half of it is \(-7\), and squaring gives \((-7)^2=49\). Therefore \(x^2-14x+49=(x-7)^2\). Option A is correct. Option B, 14, is only the magnitude of the middle coefficient, not its required half-square. Option C, 196, is \(14^2\), but the constant term uses the square of half the coefficient, not the whole coefficient. Option D, 7, is the magnitude of the half coefficient but has not been squared. The negative sign does not make the constant negative because squaring a negative number gives a positive result. A quick check is to expand \((x-7)^2=x^2-14x+49\).

(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.

A binomial square has the form \(a^2-2ab+b^2\). With \(a=2x\) and \(b=3y\), the middle term is \(-2ab=-12xy\), so A is \((2x-3y)^2\). In B, the last term should be \(9y^2\), not \(6y^2\). Exam tip: verify all three terms.

Here, \(49x^2=(7x)^2\) and \(81y^2=(9y)^2\). Using \(a^2-b^2=(a+b)(a-b)\) gives option A. Squared options produce a middle term. Exam tip: first check whether both terms are perfect squares.

With identical linear factors, the expression has the form \((x+r)^2=x^2+2rx+r^2\). Thus \(p=2r\) and \(q=r^2\), giving \(p^2=4q\). Exam tip: compare the middle coefficient with \(2r\) when spotting a perfect square.

\(9a^2-12ab+4b^2=(3a-2b)^2=(3a-2b)(3a-2b)\), so it is the product of two identical linear factors. Here \(9a^2=(3a)^2\), \(4b^2=(2b)^2\), and the middle term is \(2\times3a\times(-2b)=-12ab\). In option B, the last term is negative, whereas the square of a linear binomial has a positive last square term. Exam tip: for a perfect-square trinomial, take the square roots of the first and last terms and verify the middle term using \(\pm2pq\).

For (x^2+8x+18), the constant should be (16) for a perfect square. In exams, check the square of half the middle coefficient.