Quadratic expressions

In \(5x^2-3x\), the highest power of x is 2, so it is quadratic. A constant term is not compulsory. \(5x^3-3x\) is cubic because its highest power is 3. Exam tip: identify the degree from the highest exponent.

The expression can be recognized as a perfect square by comparing it with \((x-a)^2=x^2-2ax+a^2\). The constant term must be the square of the number inside the bracket, and the middle term must be negative twice that number times x. Checking both parts is important because the same constant can appear with either a plus or minus sign in the bracket, but the middle term selects the correct one.

Here, \(49=7^2\), so the number inside the square is 7. The middle term is \(-14x\), and \(-2\cdot7\cdot x=-14x\). Thus \(x^2-14x+49=(x-7)^2\), making option D correct. Expanding confirms it: \((x-7)^2=x^2-14x+49\). The form \((x+7)^2\) would instead have a positive middle term.

Using \((a-b)^2=a^2-2ab+b^2\), take \(a=2x\) and \(b=3\). Then \((2x-3)^2=(2x)^2-2(2x)(3)+3^2=4x^2-12x+9\). Hence, option B is correct. Option A has an incorrect coefficient of the middle term, while option D has the wrong sign. Exam tip: the middle term in \((a-b)^2\) is always \(-2ab\).

The expansion is (x^2+3x-10), so the coefficient of (x) is (3). Exam tip: the sum of constants gives the middle coefficient.

The direct answer is D: the constant term is \(-21\). Expand only as much as needed: \( (x-7)(x+3)=x^2+3x-7x-21=x^2-4x-21\). The constant term is the term without the variable, so it is \(-21\). An even quicker method is to multiply the constant numbers in the two factors: \((-7)(3)=-21\). Option A, 21, loses the negative sign. Option B, 4, is related to the coefficient of x after combining \(3x-7x\), but it is not the constant term and the coefficient is actually \(-4\). Option C, \(-4\), is the coefficient of x, not the constant term. Option D is correct because it is the product of the two constant parts. Memory cue: in a product of two linear factors, the constant term comes from multiplying the two numbers without the variable.

For a perfect square, \(x^2+bx+c=(x+b/2)^2\). Thus the constant term must be \(b^2/4\), giving \(b^2=4c\). Exam tip: equivalently, check whether the discriminant \(b^2-4c\) is zero.

The direct answer is option A: \(k=81\). To make \(x^2-18x+k\) a perfect square, compare it with \((x-a)^2=x^2-2ax+a^2\). The middle coefficient is \(-18\), so \(-2a=-18\), giving \(a=9\). Therefore the constant term must be \(a^2=9^2=81\). Indeed, \(x^2-18x+81=(x-9)^2\). Option A, 81, is correct. Option B, 36, is \(6^2\), but it would pair with a middle term \(-12x\), not \(-18x\). Option C, 18, is the magnitude of the middle coefficient, not its required half-square. Option D, 324, is \(18^2\), which is too large; the needed number is the square of half the coefficient, not the square of the whole coefficient. The standard rule is: for \(x^2+bx+c\) to be a square, \(c=(b/2)^2\), when the leading coefficient is 1. Exam cue: halve the middle coefficient first, keep its sign for the bracket, then square it; \((-9)^2\) is positive 81.

\(9x^2+12x+4=(3x)^2+2(3x)(2)+2^2=(3x+2)^2\), so it is a perfect square. In option B, the constant should be \(4\), not \(2\). Exam tip: match the middle term with \(2ab\).

Option A is \(4x^2-12x+9=(2x-3)^2=(2x-3)(2x-3)\), so it is the product of two identical binomials. Option B is the closest distractor, but \((2x-3)^2\) has constant term \(9\), not \(8\). Exam tip: for a perfect-square trinomial \(ax^2+bx+c\), check whether \(b^2=4ac\).

In \((2x-3)(x+4)\), the highest-degree term is \(2x\times x=2x^2\), so its degree is 2 and it is quadratic. Option B has degree 3. Exam tip: check for variables in denominators or radicals before calling an expression a polynomial.

The direct answer is option B: \(p=8\). Since the constant term is \(16=4^2\), a perfect-square expression with a positive middle coefficient must be \((x+4)^2\). Expanding gives \((x+4)^2=x^2+2\cdot x\cdot4+4^2=x^2+8x+16\). Comparing with \(x^2+px+16\), we get \(p=8\). Option A, 4, is the square root of the constant term, not the middle coefficient. Option B, 8, is correct because the identity doubles that root in the middle term. Option C, 16, is the constant term itself, so it is not \(p\). Option D, 32, incorrectly doubles 16 instead of doubling 4. If the coefficient were negative, the other square \((x-4)^2\) would give \(p=-8\), but the question says \(p\) is positive, so only 8 is suitable. Memory cue: for \(x^2+px+q^2\), the positive perfect-square middle coefficient is \(p=2q\).

The parts are (2x) and (5), so the middle term is (-2\cdot2x\cdot5=-20x). Exam tip: choose the negative value using (p<0).

Direct answer: option C, \(3x^2-13x-10\). Apply distribution: multiply \(3x\) by \(x\) to get \(3x^2\), and \(3x\) by \(-5\) to get \(-15x\). Then multiply \(2\) by \(x\) to get \(2x\), and \(2\) by \(-5\) to get \(-10\). The expanded expression is \(3x^2-15x+2x-10\). Combine like terms: \(-15x+2x=-13x\), giving \(3x^2-13x-10\). Option A stops before combining like terms, so it is an unsimplified intermediate step, not the requested simplified expansion. Option B has the wrong sign for \(3x\times(-5)\). Option C is correct. Option D has \(+13x\), so it has sign errors. Memory cue: use all four products and track the negative sign.

\(x^2+6x+9=x^2+2\cdot x\cdot3+3^2=(x+3)^2\), so option A is a perfect square. Option B is the closest distractor, but its constant term is \(8\); it must be \(9\) to form \((x+3)^2\). Exam tip: for \(x^2+bx+c\) to be a perfect square, check whether \(c=(b/2)^2\).

\((x+3)^2-6x=x^2+6x+9-6x=x^2+9\), whose highest power is 2, so it is quadratic. Options A and D simplify to linear expressions. Exam tip: combine like terms before identifying the degree.

In \(5x^2-9\), the highest power of x is 2, so it is quadratic. A linear x-term need not be present. \(5x-9\) is linear, while \(5x^3-9\) is cubic. Exam tip: identify the degree from the highest exponent.

The direct answer is option C: \(3x^2+5x-11\). “Subtract the second expression from the first” means \((5x^2+2x-7) - (2x^2-3x+4)\). The minus sign before the bracket changes every sign inside: \(5x^2+2x-7-2x^2+3x-4\). Now combine like terms: \(5x^2-2x^2=3x^2\), \(2x+3x=5x\), and \(-7-4=-11\). Thus the result is \(3x^2+5x-11\). Option A has \(-x\) because it fails to change the sign of \(-3x\). Option B adds the expressions instead of subtracting them. Option C performs the subtraction correctly. Option D changes several signs incorrectly and gives the wrong constant sign. Exam cue: when a whole bracket is subtracted, reverse every sign inside it before collecting like terms.

Since (9=3^2), the negative middle term is (-2\cdot3\cdot x=-6x). Exam tip: match the sign of the middle term with the condition.

The direct answer is option A. The first expression is \(x^2-4 = x^2-2^2\), so it has the form \(a^2-b^2\) and is a difference of squares. It can also factor as \((x-2)(x+2)\). The second expression is \(x^2+4x+4\). Since \(4x\) is \(2\cdot x\cdot 2\) and \(4=2^2\), it matches \(x^2+2xy+y^2=(x+y)^2\), giving \((x+2)^2\); it is a perfect square. Option A states both classifications correctly. Option B says the expressions are equal, but they are not equal for general x; for example, at \(x=1\) they are \(-3\) and \(9\). Option C reverses the two types. Option D is wrong because both expressions contain \(x^2\) and are quadratic, not linear. Memory cue: \(a^2-b^2\) has a minus sign; \(a^2+2ab+b^2\) forms a square.

In \(x^2+10x+25\), the last term is \(5^2\) and the middle term is \(2\times x\times5=10x\). Hence it equals \((x+5)^2\). In exams, verify the middle term using \(2ab\).

From (a^2=36), (a=6), and the middle term is (-2ax=-12x). Exam tip: the middle term is negative in a subtraction square.

Squaring gives (x^2-2+\frac{1}{x^2}=16), so (x^2+\frac{1}{x^2}=18). Exam tip: add (2) in the subtraction identity.

Completing the square means changing a quadratic expression into a square of a binomial plus or minus a remaining constant. For an expression beginning with \(x^2+bx\), take half of the coefficient of x, square it, and add and subtract that same value. This keeps the value unchanged while creating a perfect square.

The coefficient of x is 4. Half of 4 is 2, and its square is 4. Therefore, \(x^2+4x+7=x^2+4x+4+3=(x+2)^2+3\). The added 4 forms the square, while subtracting it from 7 leaves 3. Hence option A is correct. Expanding \((x+2)^2+3=x^2+4x+4+3=x^2+4x+7\) verifies the result.

A perfect square has the form \((x+p)^2=x^2+2px+p^2\). Thus \(b=2p\) and \(c=p^2\), giving \(b^2=4c\). Option \(b^2=c\) misses the factor 4. Exam tip: square the middle coefficient and compare it with \(4c\).

Since ((x+3)^2=x^2+6x+9), multiply by (2) and subtract (5). Exam tip: apply the outside multiplier to the whole bracket.