Introduction to Quadratic Equations

A quadratic equation has the highest power (degree) of the variable \(x\) equal to 2. In option A the highest power of \(x\) is 2 (term \(x^2\)), so it is a quadratic equation. Option B is degree 1 (linear), option C is degree 3 (cubic), and option D has no variable so it is not an equation in \(x\). Exam tip: To identify degree quickly, look for the term with the largest exponent of the variable and check that its coefficient is non-zero.

In the standard form \(ax^2+bx+c=0\), \(a\) is the coefficient of \(x^2\). In the given equation the coefficient of \(x^2\) is 5, so \(a=5\). Option B (\(-2\)) is the coefficient of \(x\) (i.e. \(b\)), option C (9) is the constant term \(c\), and option D (0) is not a coefficient present here. Exam tip: identify the number directly in front of \(x^2\) to get \(a\) quickly.

In the standard form \(ax^2+bx+c=0\), \(b\) is the coefficient of \(x\). In \(3x^2+7x-4=0\) the coefficient of \(x\) is 7, so \(b=7\). Common confusion: 3 is the coefficient of \(x^2\) (\(a\)) and -4 is the constant term (\(c\)), so those choices are incorrect. Exam tip: Always rewrite or read the equation in standard form and directly pick \(a, b, c\) before proceeding.

A quadratic is written as \(ax^2+bx+c=0\), where \(c\) is the constant term. In \(4x^2-8x+1=0\) the constant term is \(1\), so \(c=1\). The closest distractors: \(-8\) is actually \(b\), and \(4\) is \(a\); \(-1\) has the wrong sign. Exam tip: Always rewrite the equation in standard form \(ax^2+bx+c=0\) and then read off \(a, b, c\).

Write the quadratic in standard form \(ax^2+bx+c=0\). The given equation \(x^2-6x=0\) can be written as \(x^2-6x+0=0\), so the constant term is \(c=0\). Option B (\(-6\)) is the coefficient \(b\) of \(x\), not the constant; options A and D are not present in the equation. Exam tip: always convert to \(ax^2+bx+c=0\) and read off \(a, b, c\) directly.

Solve: \(9x^2+11=0\Rightarrow 9x^2=-11\Rightarrow x^2=-\dfrac{11}{9}\). Taking square roots gives complex roots \(x=\pm\sqrt{-\dfrac{11}{9}}=\pm\dfrac{i\sqrt{11}}{3}\). Option C is incorrect because it gives real roots \(\pm\dfrac{\sqrt{11}}{3}\), which do not satisfy the equation; option D is wrong because \(x=0\) yields 11, not 0. Quick exam tip: check the sign of the discriminant or move the constant to the RHS — a negative value under the square root indicates imaginary (complex) roots involving \(i\).

By definition a quadratic equation has degree 2, so the coefficient of \(x^2\) must be nonzero, i.e. \(a\neq 0\). If \(a=0\) the equation reduces to the linear form \(bx+c=0\). Options C and D (\(b=0\) or \(c=0\)) are only special cases but do not prevent the equation from being quadratic as long as \(a\neq 0\). Exam tip: always check the coefficient of \(x^2\) to decide if an equation is quadratic.

Standard form means writing the quadratic as \(ax^2+bx+c=0\) with all terms on one side. Moving the 3 from the right to the left changes its sign to −3, giving \(2x^2+5x-3=0\). Option A is wrong because it keeps the constant as +3 instead of −3. Options C and D are wrong because they incorrectly change the sign of the linear term 5x. Exam tip: When converting to standard form, move all terms to one side and remember to change the sign of any term you transpose; then compare with \(ax^2+bx+c=0\).

The standard form places all terms on one side and zero on the other. Moving the terms from the right to the left changes their signs, giving \(x^2 - 4x - 12 = 0\). The other options fail because their signs are not changed correctly — e.g. option C has the constant sign wrong, options B and D have the linear term sign wrong. Exam tip: Bring all terms to one side and arrange in descending powers of \(x\) (\(x^2\), then \(x\), then constant) before finalizing the answer.

The standard form \(ax^2+bx+c=0\) requires the right-hand side to be 0 and the coefficient \(a\) of \(x^2\) to be nonzero. Option A is already in that form (a=1, b=2, c=1). Options B and C have nonzero terms on the right as written (they can be rearranged into standard form, e.g. B → \(x^2+2x-1=0\), but the question asks which is in standard form as given). Option D lacks an \(x^2\) term so it is not a quadratic equation. Exam tip: verify RHS = 0 and that the \(x^2\) coefficient is not zero to confirm standard form quickly.

Substitute and evaluate: \(2^2-5\cdot2+6=4-10+6=0\). Hence the left-hand side equals 0 and \(x=2\) is a root of the quadratic. Option C (4) is incorrect — that would result from using only \(2^2\) and ignoring the other terms. Exam tip: compute powers and multiplications first, then perform additions/subtractions to avoid sign or order errors.

Check by substitution: \(3^2-9=9-9=0\), so 3 is a root. Factoring gives \(x^2-9=(x-3)(x+3)\), so the roots are \(x=3\) and \(x=-3\) (note -3 is not among the choices). The nearest distractor 2 fails because \(2^2-9=4-9=-5\neq0\). Exam tip: for simple quadratics either factor or test small integer values to verify roots quickly.

Substituting \(x=2\) gives \(2^2-4\cdot2+4=4-8+4=0\). So the equation becomes \(0=0\), which means \(x=2\) is a correct solution (root). Also note the polynomial is \((x-2)^2\), so this is a repeated root (discriminant zero). Exam tip: always compute each term carefully when substituting; recognizing a perfect square trinomial saves time.

A quadratic equation has highest power 2, so its standard form is \(ax^2+bx+c=0\), with \(a\ne0\). Option B is linear because its highest power is 1. Exam tip: identify the equation type by checking the highest exponent.

The degree of a polynomial is the highest exponent of the variable; a quadratic equation must have degree 2. Option C has highest exponent \(3\) (\(x^3+x+1=0\)), so it is not quadratic. The other options are degree 2: A is degree \(2\), B is degree \(2\), and D becomes \(4x^2-7=0\) which is also degree \(2\). The closest distractor is D because it is written as \(4x^2=7\) and may look different, but converting to standard form shows it is quadratic. Exam tip: convert to standard form \(ax^2+bx+c=0\) and check the highest power of the variable.

Write the quadratic in standard form \(ax^2+bx+c=0\). The equation \(6x^2=0\) is equivalent to \(6x^2+0x+0=0\), so \(a=6,\; b=0,\; c=0\). Option C is wrong because it incorrectly gives the x-coefficient as 1 instead of 0; B and D place numbers in wrong positions. Exam tip: explicitly add missing terms with coefficient 0 and then read off a, b, c from the standard form.

Expanding gives \\x(x+5)=x\cdot x + x\cdot 5 = x^2+5x\\. In standard form this is \\x^2+5x+0=0\\ or simply \\x^2+5x=0\\. Option B has the wrong sign (would be \\-5x\\ if the second term were negative), option C incorrectly multiplies the square term by 5 (\\5x^2\\ instead of \\x^2\\), and option D is linear, not quadratic. Exam tip: expand brackets carefully and collect terms on one side to match \\ax^2+bx+c=0\\ before identifying a, b, c.

Multiply the two binomials using FOIL: \((x+2)(x+3)=x\cdot x + x\cdot 3 + 2\cdot x + 2\cdot 3 = x^2+3x+2x+6\). Combining like terms gives \(x^2+5x+6=0\), so A is correct. The closest distractor B (\(x^2+6x+5=0\)) simply swaps coefficients and constant but does not follow from multiplication; D has the wrong sign on the constant. Exam tip: use FOIL (or multiply each term) and then combine like terms to avoid sign or order mistakes.

Expanding gives \((x-4)(x+1)=x^2+x-4x-4=x^2-3x-4\), so the standard form is \(x^2-3x-4=0\). Quick check: the roots are 4 and −1, their sum is 3, and since −b/a = 3 we must have b = −3. The closest distractor (option B) typically arises from a sign mistake on the linear term; the other options have wrong signs or constant term. Exam tip: combine like terms immediately after multiplying and verify using sum/product of roots.

Direct answer: Option B, 1. A quadratic equation is commonly written as \\(ax^2+bx+c=0\\), where a is the coefficient of \\(x^2\\). In the expression \\(x^2+8x+16=0\\), the first term can be written as \\(1x^2\\). Therefore a=1. Option A, 0, is wrong because the term is present, not absent. Option B is correct because an unwritten numerical factor is understood to be one. Option C, 8, is the coefficient of x, not of \\(x^2\\). Option D, 16, is the constant term because it has no x. Read the power carefully: the coefficient belongs to the term with that exact power. Exam cue: an alone-standing variable term has coefficient 1.

In the standard form \(ax^2+bx+c=0\), \(a\) is the coefficient of \(x^2\). In \(-x^2+3x+10=0\) the \(x^2\) term is \(-x^2\), so its coefficient is \(-1\). Therefore \(a=-1\). Note: 3 is the coefficient \(b\) of \(x\), 10 is the constant \(c\), and 1 is the wrong sign. Exam tip: rewrite any quadratic as \(ax^2+bx+c=0\) and compare coefficients term by term.

\(x^2+6x+9\) is a perfect-square trinomial. Comparing with \((x+a)^2=x^2+2ax+a^2\) gives \(2a=6\), so \(a=3\) and the trinomial equals \((x+3)^2\). Hence the equation factors as \((x+3)^2=0\). Option B is incorrect because \((x-3)^2\) would produce a middle term of \(-6x\); options C and D have wrong constant terms. Exam tip: identify perfect squares by checking if the constant equals the square of half the middle coefficient (i.e., check if constant = (middle/2)^2).

Factor the quadratic: \(x^2+x-6=(x+3)(x-2)\). Thus the roots are \(x=-3\) and \(x=2\). The value 2 is among the options, so it is the correct answer. Checking the nearest distractor 1 gives \(1^2+1-6=-4\), not zero, so it is incorrect. Exam tip: either factor the quadratic or substitute the option values to verify quickly.

The governing definition is the degree of a polynomial equation. The degree is the greatest exponent of the variable that has a non-zero coefficient. When the highest power of x is 2, the equation is called a quadratic equation. Its usual standard form is ax² + bx + c = 0, where a is not zero; this condition ensures that the x² term is actually present. A linear equation has degree 1, a cubic equation has degree 3, and a constant equation has degree 0 because it contains no variable term of positive degree. Therefore, an equation of degree 2 is precisely a quadratic equation, making option B correct. The other choices refer to different degrees.

A quadratic (second-degree) expression must include a nonzero \(x^2\) term. Option C, \(x^2-2=0\), clearly contains the \(x^2\) term so it is correct. The other options, \(5x+2=0\), \(7x-9=0\), and \(3x+4=7\), are linear equations whose highest power of \(x\) is 1 and therefore do not contain \(x^2\). Exam tip: To identify a quadratic quickly, check the highest power of the variable — if it is 2, the equation is quadratic.