Introduction to Quadratic Equations

Here the coefficient of (x^2) is (0), so the (x^2) term disappears. For a quadratic equation, (a\neq 0) is necessary.

Multiply each term of the first binomial by each term of the second: \((x+1)(x+2)=x\cdot x + x\cdot 2 + 1\cdot x + 1\cdot 2 = x^2+2x+x+2 = x^2+3x+2\). Thus the expanded equation is \(x^2+3x+2=0\). The closest distractor \(x^2+3x-2=0\) has the constant term sign wrong; the other options have incorrect middle or constant terms. Exam tip: use FOIL and combine like terms carefully; check signs for the constant term.

Standard form means writing all terms on one side as \(ax^2+bx+c=0\). For \(x^2=7\), move 7 to the left to get \(x^2-7=0\), which is the standard form. Option A is incorrect because \(x^2+7=0\) corresponds to \(x^2=-7\). Option C is wrong since it introduces a \( -7x\) term not present in the original equation. Exam tip: To convert to standard form, bring every term to the left and arrange as \(ax^2+bx+c=0\).

This is a difference of squares: \(x^2-1=(x-1)(x+1)\). The equation equals zero when \(x-1=0\) or \(x+1=0\), so \(x=1\) and \(x=-1\). Option A (0,1) is incorrect because at \(x=0\) we get \(0^2-1=-1\neq0\). Options C and D also do not satisfy the equation. Exam tip: recognize \(x^2-a^2\) and factor, or take square roots to write \(x=\pm a\).

From \(x^2=9\) we get \(x=\pm\sqrt{9}=\pm3\), so the roots are \(-3\) and \(3\). Option B wrongly lists 9 (although 3 is correct) — but \(9^2=81\), not 9. Option D has ±9 whose squares are 81, and option A contains 0 whose square is 0, so none of those satisfy the equation. Exam tip: for equations of the form \(x^2=a\) always take both ±√a and verify by substituting back.

For roots (0) and (-3), the factors are (x) and (x+3). So the equation is (x(x+3)=0), that is (x^2+3x=0).

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.

Comparing with standard form, the constant term is absent, so (c=0). Treat a missing constant term as (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.

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).