Introduction to Quadratic Equations

Correct answer: A. Yes

To check a root substitute the value into the equation: \(1^2-1=1-1=0\). Thus \(x=1\) satisfies the equation and is a root. The closest distractor "Only \(x=0\) is a root" is incorrect because factoring gives \(x(x-1)=0\), so both \(x=0\) and \(x=1\) are roots. Exam tip: Either substitute the candidate value or factor the quadratic to identify roots quickly.

Correct answer: C. Quadratic polynomial

(p(x)=x^2+2x+1) is a polynomial because it is not set equal to (0). It becomes an equation when written as equal to (0).

Correct answer: B. \(2\)

The degree of a polynomial equation is the highest power of \(x\) present in it. In \(x^2+1=0\), the highest power of \(x\) is \(2\) (from the term \(x^2\)). So the degree is \(2\).

Correct answer: A. -1

In the equation \(2x^2 - x + 8 = 0\) the linear term is \(-x\). The coefficient is the number multiplied by \(x\), so here it is \(-1\). Hence the correct answer is \(-1\). Option B (1) is incorrect because it ignores the negative sign; options C and D are unrelated values. Exam tip: First identify the linear term and always include its sign when stating the coefficient.

Correct answer: D. (0)

The correct answer is option D, c=0. A quadratic equation is commonly written as ax^2+bx+c=0, where c is the constant term, the term without x. In x^2+7x=0, the coefficient of x^2 is 1, the coefficient of x is 7, and there is no separate constant term. A missing term has coefficient zero, so the equation can be written as x^2+7x+0=0. Therefore c=0. Option A, 1, is the coefficient of x^2, so it is a, not c. Option B, 7, is the coefficient of x, so it is b, not c. Option C, −7, is not present as any coefficient in the given equation. Option D correctly represents the absent constant term. A helpful check is to compare each term with its position in ax^2+bx+c=0: x^2 gives a, x gives b, and the number alone gives c.

Correct answer: B. \(x^2-5x=0\)

Expanding gives \(x(x-5)=x\cdot x - x\cdot 5 = x^2-5x\), so the standard form is \(x^2-5x=0\). Option A has the wrong sign (should be -5x, not +5x); option C has incorrect coefficients; option D incorrectly changes the linear term to a constant. Exam tip: apply distributive property term-by-term and pay attention to signs when removing brackets.

Correct answer: B. \(x^2-6x=0\)

Bring the term on the right, \(6x\), to the left; the sign changes giving \(x^2-6x=0\). The standard form is \(ax^2+bx+c=0\) with all terms on one side. Option A has the wrong sign for the \(x\)-term; options C and D do not represent the same rearrangement of the original equation. Exam tip: always move all terms to one side so the equation equals zero and then identify \(a, b, c\).

Correct answer: C. 3x² + 1 = 0

The degree of an equation in one variable is determined by the greatest exponent of that variable. In option A, the degree is 1; in option B, it is 3; in option C, the greatest exponent is 2; and option D contains a square root, equivalent to x^(1/2). Therefore, 3x² + 1 = 0 has highest power 2 and is a quadratic equation. Hence option C is correct.

Correct answer: A. Because it has a negative power of (x)

(\frac{1}{x^2}=x^{-2}), which is not polynomial form. A quadratic equation cannot have a negative power of the variable.

Correct answer: D. Because it has a (\sqrt{x}) term

The term (\sqrt{x}) shows a fractional power of the variable, so it is not in usual quadratic form. A quadratic equation has only (x^2), (x), and constant terms.

Correct answer: C. \(x^2 + 2x + 1 = 0\)

Expanding \((x+1)^2\) gives \(x^2 + 2x + 1\). The standard quadratic form places all terms on one side equal to zero, so the correct form is \(x^2 + 2x + 1 = 0\). Option B is incorrect because \(x^2 - 2x + 1\) is the expansion of \((x-1)^2\); options A and D omit or alter the linear term. Exam tip: use the identity \((a+b)^2 = a^2 + 2ab + b^2\) to expand quickly and then set the expression equal to zero.

Correct answer: B. -3

Substitute the root \(x=2\) into the polynomial: \(2^2+k\cdot2+2=0\) so \(4+2k+2=0\). Thus \(2k=-6\) and \(k=-3\). Option A (3) is incorrect because substituting 3 gives \(4+6+2=12\neq0\). Option D (\(-2\)) is also wrong since it yields \(4-4+2=2\neq0\). Exam tip: use direct substitution or the Factor Theorem — if \(x=a\) is a root then the polynomial evaluated at \(a\) equals zero.

Correct answer: A. \(3x^2-12=0\)

Use the standard form \(ax^2+bx+c=0\). Substituting gives \(3x^2+0x-12=0\), and since the \(x\)-term is absent this simplifies to \(3x^2-12=0\). You can further divide by 3 to get \(x^2-4=0\). Closest distractors fail for clear reasons: option C wrongly includes a \(-12x\) term (but \(b=0\)), option B has the wrong sign on the constant, and option D swaps coefficients. Exam tip: always substitute coefficients carefully and then simplify; check whether the \(x\)-term should be present when \(b=0\).

Correct answer: C. Linear term

There is no (x) term, so the linear term is absent. The coefficient of a missing term is considered (0).

Correct answer: A. (2y^2-3y+1=0)

The highest power of (y) is (2), so it is quadratic in (y). The variable may be different, but the degree must be (2).

Correct answer: B. a

In the standard form \(ax^2+bx+c=0\), the coefficient of \(x^2\) is \(a\). The coefficient is the number multiplying that power of \(x\); if \(a=0\) the equation would not be quadratic. Option A (\(b\)) is incorrect because \(b\) is the coefficient of \(x\); option C (\(c\)) is the constant term; option D (\(x\)) is the variable itself. Exam tip: always check that the leading coefficient \(a\) is nonzero when identifying a quadratic equation.

Correct answer: A. Yes because the coefficient of (x^2) is (1)

In (x^2+4=0), the coefficient of (x^2) is (1), so it is quadratic. Having real roots is not a condition for being quadratic.

Correct answer: B. 3

The expression \(ax^2+bx+c\) contains three possible terms: \(ax^2\), \(bx\), and \(c\). Therefore the maximum number of different terms is \(3\).

Correct answer: D. Yes because the coefficient of (x^2) is (2)

In (2x^2=0), the coefficient of (x^2) is (2\neq 0). It can be quadratic even without linear and constant terms.

Correct answer: A. (x^2+3x-10=0)

The area is (x(x+3)=10), which gives (x^2+3x-10=0). In geometry questions, form the equation using the formula.

Correct answer: B. −2x² + 9x − 7 = 0

A quadratic equation in standard form is ax² + bx + c = 0. The values supplied are a = −2, b = 9 and c = −7. Substituting them in their respective positions gives (−2)x² + 9x + (−7) = 0, which simplifies to −2x² + 9x − 7 = 0. Therefore option B is correct. The coefficient a belongs to x², b belongs to x, and c is the constant term; changing their positions produces a different equation. Option A loses the negative sign of a, option C changes the sign of b, and option D swaps the roles of a and c. Keeping the sign and position of every coefficient is essential.