Introduction to Quadratic Equations

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.

(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).

The degree of a polynomial is the highest exponent of the variable with a nonzero coefficient. The left-hand side polynomial x^2+1 has highest power 2, so the degree is 2. Option A (1) is incorrect because that would require a highest power of x to be 1; option C (0) would apply only to a nonzero constant polynomial; option D (4) is wrong since there is no x^4 term. Exam tip: To find the degree of an equation, look at the polynomial on the left side in standard form and ignore constant terms—take the largest exponent of the variable.

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.

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.

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.

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 \(a, b, c\).

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.

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

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.

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.

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.

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 \(b=0\).

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

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.

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.

A quadratic equation has highest power 2, so its standard form is written as \\(ax^2+bx+c=0\\), where \\(a\\ne0\\). This form can contain three kinds of terms: the quadratic term \\(ax^2\\), the linear term \\(bx\\), and the constant term \\(c\\). If a coefficient is zero, one of these terms may disappear in a particular equation, but the maximum possible number of different terms remains three.

For example, \\(2x^2+5x+7=0\\) contains all three terms. The symbols \\(a\\), \\(b\\), and \\(c\\) are coefficients or constants attached to these terms, not extra terms themselves. Hence the maximum is 3, which is option C. Option 2 would apply only when one term is absent, while option 4 incorrectly counts something that is not a separate algebraic term.

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

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

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.