Quadratic expressions

Expanding \((2x+5)(x+2)\) gives \(2x^2+4x+5x+10\). Therefore, the middle term is \(4x+5x=9x\). Comparing it with \(2x^2+mx+10\), we get \(m=9\). The values 4 and 5 are coefficients of the separate cross terms, whereas \(m\) is their sum. Exam tip: add both cross terms when finding the middle term in a product of two binomials.

(81=9^2), so the negative middle term is (-2\cdot x\cdot9=-18x). In exams do not forget the given sign.

The zeroes are \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Two distinct real zeroes require a positive quantity inside the square root, so \(b^2-4ac>0\). If it equals zero, the two zeroes coincide. Exam tip: check the discriminant first to identify the nature of zeroes.

(x^2+6x+9=(x+3)^2), so (11=9+2). In exams form the needed square and keep the remainder separate.

Since \(25=5^2\), we have \(x^2-25=x^2-5^2\). Using \(a^2-b^2=(a-b)(a+b)\), it factorises as \((x-5)(x+5)\). Exam tip: check whether both terms are squares.

Every variable term in \(3x^2-2xy+y^2+5\) has total degree 2; for example, the degree of \(xy\) is \(1+1=2\). Hence A is quadratic. In B, each term has degree 3. Exam tip: add exponents in each term to identify its degree.

(a^2+b^2=(a-b)^2+2ab=25+12=37). In exams add (2ab) when using the square of difference identity.

In \((x+p)(x-p)=x^2-p^2\), the middle terms cancel. Hence \(x^2-81=x^2-9^2=(x+9)(x-9)\). \(x^2+18x+81\) is a perfect square, not a product of conjugate binomials. Exam tip: opposite constants give a zero middle term.

On simplifying option B, \((x+3)(x-3)+9=x^2-9+9=x^2\). Its highest power is 2, so it is a quadratic expression. Options A and D simplify to 4 and 1 respectively, while option C simplifies to \(5x\), which is linear. Exam tip: simplify an expression completely before identifying its degree.

A quadratic expression must have degree 2, so the coefficient of \(x^2\), namely \(a\), must be non-zero. The coefficients \(b\) and \(c\) may be zero. Exam tip: check that the highest-power coefficient is non-zero first.

The cross terms (10x) and (3x) combine to (13x). In exams check options using both leading coefficient and constant term.

\(4x^2-25y^2=(2x)^2-(5y)^2\). Using \(a^2-b^2=(a+b)(a-b)\), it becomes \((2x+5y)(2x-5y)\). Options C and D are perfect squares; first check for a difference of squares.

Let \(x^2+bx+c=(x+a)^2\). Since \((x+a)^2=x^2+2ax+a^2\), we get \(b=2a\) and \(c=a^2\). Hence \(b^2=4a^2=4c\). In exams, square the middle coefficient and compare it with \(4c\).

In standard form, the (x^2) term comes first, then the (x) term, then the constant term. In exams arrange terms in descending powers.

In option A, \(9x^2=(3x)^2\) and \(16y^2=(4y)^2\). The required middle term for these squares is \(-2(3x)(4y)=-24xy\). Hence, \(9x^2-24xy+16y^2=(3x-4y)^2\), so it is the perfect square of a binomial. In option B, the last term is \(12y^2\), not \(16y^2\), so it cannot form this perfect square. Exam tip: take the square roots of the first and last terms, then check whether the middle term equals \(\pm2ab\).

((x+5)(x+7)=x^2+12x+35), so (s=12). In exams the sum of constant parts is the middle coefficient.

(64=8^2) and (-16x=-2\cdot x\cdot8). In exams check both the last term and the middle term in a perfect square.

The number attached to (x^2) is (3). In exams, identify a coefficient from the numerical part attached to the term.

In standard form, write the (x^2) term, then the (x) term, then the constant. In exams, follow descending powers.

When \(a\ne0\), the highest power of \(x\) is 2, so the expression is quadratic. \(b\) and \(c\) may be zero; if \(a=0\), it becomes linear or constant. Exam tip: check the non-zero coefficient of the highest power first.