Introduction to Quadratic Equations

A quadratic must be in standard form \(ax^2+bx+c=0\). Rewrite \(3x^2-21=0\) as \(3x^2+0x-21=0\). Hence \(a=3,\; b=0,\; c=-21\). The most tempting wrong choice is B (3, -21, 0) — it incorrectly swaps the linear and constant coefficients; the linear coefficient \(b\) is zero because there is no \(x\) term. Exam tip: always convert the equation to \(ax^2+bx+c=0\) first so you can read off \(a,b,c\) directly.

A quadratic equation in a variable means the highest power (degree) of that variable is 2. In option A the variable \(t\) appears as \(t^2\), so its highest power is 2 and the equation is quadratic in \(t\). Option B is quadratic but in \(x\), not in \(t\), so it is not correct for this question. Options C and D are linear in \(t\) (highest power 1). Exam tip: always check the highest exponent of the specific variable named in the question, not just whether any variable is squared.

Expand the factor: \(4x(x-3)=4x\cdot x-4x\cdot3=4x^2-12x\). Thus the standard quadratic form (ax^2+bx+c=0) is \(4x^2-12x=0\). Option B is wrong because the sign is incorrect (+12x instead of -12x). Option C is wrong because the coefficient 4 is missing, and option D is linear (not quadratic). Exam tip: always expand factors first and collect like terms to write the expression as \(ax^2+bx+c=0\).

Compute \(p(2)=2^2-4=4-4=0\). Thus the value is 0 and \(x=2\) is a root of the polynomial. The closest distractor (option C = 4) arises if a student computes only \(2^2\) and forgets to subtract 4. Exam tip: when evaluating polynomials, substitute the number and follow order of operations—evaluate powers first, then perform addition/subtraction.

The constant term is the term that does not contain the variable x (degree zero). 6x^2 and x contain x, so they are not constant. The only term without x is −8, so the constant term is −8. Exam tip: substitute x = 0 into the polynomial; the result equals the constant term (here 6·0^2 + 0 − 8 = −8).

Write the equation in the standard form \(ax^2+bx+c=0\): \(x^2-13x+0=0\). The coefficients are \(a=1, b=-13, c=0\). The constant term is \(c=0\), so the constant term is absent. The closest distractor (linear term) is wrong because the term \(-13x\) is present. Exam tip: always rewrite as \(ax^2+bx+c\) and read off \(a,b,c\).

A pure (or 'pure') quadratic has the form ax^2+bx+c=0 with a ≠ 0 and the linear coefficient b = 0. In 4x^2-16=0 the x-term is absent (b=0) and a=4 ≠ 0, so it is a pure quadratic. x^2+5x+6=0 has a nonzero linear term (5x) so it is not pure. x+16=0 is linear (degree 1) and x^3-16=0 is cubic (degree 3), so neither is a quadratic. Exam tip: check the coefficient of x quickly — if it is zero and the x^2 coefficient is nonzero, the equation is a pure quadratic.

Substituting \(x=-3\) gives left-hand side = \((-3)^2+6(-3)+9=9-18+9=0\). Hence the equation holds and \(x=-3\) is a solution. Option B is incorrect because the equation is satisfied; options C and D are incorrect since the left-hand side evaluates to 0, not 6 or −6. Exam tip: when substituting negative numbers, keep them in parentheses and simplify step by step to avoid sign errors.

Using the standard form \(ax^2+bx+c=0\) and substituting \(a=3,b=-4,c=-2\) yields \(3x^2-4x-2=0\), so A is correct. The closest distractor B has the sign of \(b\) reversed (it shows \(+4x\)), making it incorrect. Options C and D change the positions or values of the coefficients. Exam tip: always substitute coefficients into \(ax^2+bx+c\) and carefully note the sign of \(b\).

The standard form is \(ax^2+bx+c=0\). Substituting a=2, b=0, c=-18 gives \(2x^2+0x-18=0\), which simplifies to \(2x^2-18=0\). Option B is wrong because the sign of c is reversed (+18 instead of -18). Option C is incorrect because it includes an x-term even though b=0. Option D swaps coefficients (incorrect values for a and c). Exam tip: always write \(ax^2+bx+c=0\) first, substitute the coefficients, then remove any zero terms.

Standard form for a quadratic is \(ax^2+bx+c=0\); to convert, bring the RHS term(s) to the left and change their signs. In option A, moving 8 to the left gives \(-8\), so \(x^2+2x=8\) becomes \(x^2+2x-8=0\) — correct. Option B is wrong because \(x^2+2x=-8\) would become \(x^2+2x+8=0\). Options C and D are incorrect because they have \(-2x\) (different sign for the linear term) not \(+2x\). Exam tip: always move all terms to one side to match \(ax^2+bx+c=0\) before comparing coefficients.

Core idea: In a quadratic of the form \(ax^2+bx+c=0\), \(a,b,c\) are the coefficients of \(x^2, x\) and the constant term respectively. Here \(a=5,\; b=5,\; c=5\). Thus \(a+b+c=5+5+5=15\). Option B (10) is incorrect because it corresponds to adding only two terms rather than all three coefficients. Exam tip: Always identify the coefficients from the standard form \(ax^2+bx+c\); include the constant term as \(c\).

In standard form \(ax^2+bx+c=0\) we have \(a=4,\; b=-2,\; c=-6\). So \(a-b+c=4-(-2)+(-6)=4+2-6=0\). Option B (2) typically arises from a sign mistake — for example, if one incorrectly takes \(b=+2\) then \(4-2-6=-4\), not 2. Exam tip: always rewrite the equation as \(ax^2+bx+c=0\) and read off the signs of \(a,b,c\) carefully before substituting.

The highest power (exponent) of the variable x is \(2\) (from the term \(x^2\)), so the degree is 2 and the equation is quadratic. Option D (6) is incorrect because 6 is the coefficient of \(x^2\), not the exponent; do not confuse coefficient with degree. Exam tip: The degree of a polynomial is the largest exponent of the variable, not the largest numerical coefficient.

Expand \\((x+4)(x-3)\\): \\(x^2 -3x +4x -12 = x^2 + x -12\\), so it matches the given quadratic. Closest distractor B expands to \\((x-4)(x+3)= x^2 - x -12\\), which has the middle term \\(-x\\) (wrong sign). Options C and D expand to \\(x^2+7x+12\\) and \\(x^2-7x+12\\) respectively, so they do not match. Exam tip: find two numbers whose product is \\-12\\ and sum is \\+1\\ (here 4 and -3).

Area of a square equals the square of its side. With side \\(x+2)\\ the area is \\(x+2)^2\\, and since the area is 49 the equation is \\( (x+2)^2 = 49)\\. Expanding gives \\x^2+4x+4=49\\ and rearranging yields the quadratic \\x^2+4x-45=0\\. Option B (\\x+2=49\\) is incorrect because it treats the side as if it were the area. Options C and D are wrong because they show incorrect algebraic manipulation (missing the square or misplacing terms). Exam tip: directly apply Area = (side)^2 and then bring terms to one side to get the standard quadratic form before solving.

Consecutive even numbers differ by 2, so if the smaller is x the next is x+2. The product equation is therefore \(x(x+2)=48\). Option B is incorrect because \(x+1\) represents consecutive integers, not consecutive even numbers. Option C is wrong because it gives a sum, not a product. Option D is incorrect because \(2x=48\) does not represent the product of two different consecutive even numbers. Exam tip: explicitly write the two numbers (x and x+2) first, then form the product or sum as stated in the problem.

The standard form of a quadratic equation is usually written as ax^2 + bx + c = 0. Starting with x^2 = 6x + 7, move every term on the right to the left side. When a term crosses the equality sign, its sign changes: 6x becomes -6x and 7 becomes -7. Therefore, the correct standard form is x^2 - 6x - 7 = 0.

Option A is this correct form. Option B writes x^2 + 6x + 7 = 0, which changes neither sign and is not equivalent to the original equation. Option C, x^2 - 6x = 7, is an equivalent rearrangement, but it is not yet the usual standard form because the constant has not been moved to the left. Option D simply repeats the original equation. Hence option B shows the error.

Expand the perfect square: \((x+3)^2 = x^2+6x+9\). Subtract 25 from both sides: \(9-25=-16\), giving \(x^2+6x-16=0\). Thus option A is correct. Option B has the wrong sign on the constant (+16 instead of -16). Options C and D have incorrect coefficients for x or the constant term. Exam tip: always expand the square first, move all terms to one side and then combine like terms to reach the standard form.

An equation is quadratic only when the coefficient of \(x^2\) is nonzero. In \((k-1)x^2+3x+2=0\), that coefficient is \(k-1\). It must therefore satisfy \(k-1\ne0\). Solving this condition gives \(k\ne1\), so option B is correct.

If \(k=1\), the quadratic term disappears and the equation becomes \(3x+2=0\), which is linear rather than quadratic. Any other value of \(k\) leaves a nonzero coefficient on \(x^2\), so the equation remains quadratic. The values \(k=0\) and \(k=-1\) are allowed examples, but they are not the complete condition. Thus the supplied answer correctly states the necessary and sufficient condition.

By definition a quadratic equation has degree \\(2\\) and is written in standard form \\(ax^2+bx+c=0\\) with the necessary condition \\(a\ne0\\). The constant term \\(c\\) can be zero (e.g. \\(ax^2+bx=0\\) or \\(ax^2=0\\)), so it is not required that \\(c\ne0\\). Hence option D is false. The common misconception is thinking every coefficient must be nonzero; that would make it not generally quadratic. Exam tip: verify statements by substituting a simple example such as \\(x^2+2x=0\\).

Multiply term-by-term: 2x·x = 2x^2, 2x·(−4) = −8x, 1·x = x, 1·(−4) = −4. Adding gives 2x^2 −8x + x −4 = 2x^2 −7x −4, so option A is correct. Closest distractors are wrong due to sign or constant errors (B has +7x instead of −7x; D has +4 instead of −4). Exam tip: multiply each term carefully and combine like terms; verify the constant by multiplying the constant terms of the binomials.

Substitute \(x=-1\) into the equation: \((-1)^2 + m(-1) - 6 = 0\). This gives \(1 - m - 6 = 0\), i.e. \(-m - 5 = 0\). Therefore \(m = -5\). The option 5 arises from forgetting the negative sign on \(m\); if \(m=5\) the equation is not satisfied. Exam tip: Always substitute the root directly and simplify carefully, paying attention to signs and powers.

A quadratic equation has highest power (2) of (x) and (a\neq 0). In exams identify the standard form first.

A quadratic equation is a polynomial equation whose highest power (degree) of the variable is 2. In option A the highest power is 2, so it is quadratic. Option B is linear (degree 1). Option C is cubic (degree 3). Option D is not a polynomial because the variable appears in the denominator, so it cannot be a quadratic. Exam tip: first check the highest exponent and ensure the expression is a polynomial (no variables in denominators).