Introduction to Quadratic Equations

Answer: B, 2. The degree of a polynomial equation is the greatest exponent of the variable whose coefficient is not zero. In 7x² − 4x + 9 = 0, the terms are 7x², −4x, and 9. Their powers of x are 2, 1, and 0 respectively. The coefficient of x² is 7, which is non-zero, so the highest actual power is 2. Therefore the equation has degree 2 and is called a quadratic equation. A, 1, describes only the linear term −4x and ignores the higher-power term. C, 3, is not present anywhere in the equation, so it cannot be the degree. D, 0, is the degree of a non-zero constant polynomial, not of this expression. The equals-zero sign does not change the degree; we inspect the polynomial on the left. Memory cue: degree means the largest exponent that remains after ignoring zero-coefficient terms.

Compare with the standard form \(ax^2+bx+c=0\). Here the coefficient of \(x^2\) is 1 so \(a=1\); the coefficient of \(x\) is -5 so \(b=-5\); the constant term is 6 so \(c=6\). Option B is a common sign-mistake (it shows \(b=5\) instead of \(-5\)). Option C incorrectly negates \(a\), and D has the terms permuted. Exam tip: always identify the coefficient of \(x^2\) first and copy signs exactly from the equation.

A quadratic must have a nonzero coefficient for the highest power \(x^2\). If \(a=0\), the \(x^2\) term disappears and the equation reduces to the linear form \(bx+c=0\). Thus the required condition is \(a\neq 0\). Option B (\(a<0\)) is incorrect because \(a\) need not be negative; it can be positive. Option D (\(a=1\)) is only a special case, not a necessity. Exam tip: to confirm the degree of a polynomial, always check that the coefficient of the highest-power term is nonzero.

Substituting \(x=2\) gives \(2^2-4=4-4=0\), so \(x=2\) is indeed a root. Option C is incorrect because \(x=-2\) also satisfies the equation (\((-2)^2-4=0\)), so it is false to claim 'only \(x=-2\)' is a root — both \(2\) and \(-2\) are roots. Exam tip: To check a candidate root, substitute it into the polynomial and verify the result equals \(0\).

Square means (x^2) and (3) times means (3x). In word problems, convert the sentence slowly into an equation.

The governing concept is the degree of an equation. A quadratic equation must contain a variable whose highest power is exactly 2, and the coefficient of that squared term must be non-zero. In 3x + 2 = 0, the variable x occurs only as x¹, so the highest power is 1 and the equation has degree 1. It is therefore a linear equation. The constant term 2 does not make an equation quadratic; linear equations can also contain constant terms. Option A is incorrect because there is no x² term at all. Option C is incorrect because x is clearly present. Hence option D correctly explains why the equation is not quadratic.

A quadratic equation has degree (2), so it can have at most (2) roots. Remember degree and maximum number of roots together.

The governing concept is identification of coefficients in a polynomial or quadratic expression. A coefficient is the numerical factor that multiplies a variable term. In 5x² − 2x + 1 = 0, the term containing x² is 5x², so the coefficient of x² is 5. The negative sign before 2 belongs to the term −2x and therefore describes the coefficient of x, not the coefficient of x². The number 1 is the constant term, so it cannot be the requested coefficient. Option B ignores the sign and refers to the x-term, option C is the constant, and option D is the coefficient of x. Therefore option A is the only correct answer.

To write in standard form bring all terms to one side as \(ax^2+bx+c=0\). From \(4 = 3x - x^2\), moving the right-hand terms to the left changes signs: \(-x^2\) becomes \(+x^2\), \(+3x\) becomes \(-3x\), and \(+4\) stays. So the standard form is \(x^2-3x+4=0\). The closest distractor is option C (\(x^2-3x-4=0\)) which has the constant term sign wrong; option B has the sign of the linear term wrong. Exam tip: always rearrange to \(ax^2+bx+c=0\) and check sign changes when moving terms across ‘=’.

There is no (x) term, so the linear term is considered (0x). A missing term has coefficient (0).

The constant term is the term that does not contain the variable x. In \(2x^2+3x-8\) the only term without x is \(-8\), so the constant term is \(-8\). Option C (8) is a common distractor but incorrect because the sign is negative. Exam tip: to find the constant quickly, scan for the term with no x and pay attention to its sign.

A quadratic term is the term in which the variable has exponent 2 (i.e. contains \(x^2\)). In the equation \(6x^2 - x + 5 = 0\), only \(6x^2\) contains \(x^2\), so it is the quadratic term. The distractors \(-x\) and \(x\) are linear terms (exponent 1) and \(5\) is a constant (exponent 0). Exam tip: check the exponent of the variable — the term with \(x^2\) is the quadratic term.

A pure quadratic equation has the form \(ax^2+bx+c=0\) with the linear coefficient \(b=0\), so it reduces to \(ax^2+c=0\). In \(4x^2-9=0\) there is no \(x\) term, so it is a pure quadratic. The closest distractor, \(x^2+2x+1=0\), is a quadratic but not pure because of the \(2x\) term. Exam tip: check the coefficient of \(x\); if it is zero (or the \(x\) term is absent) the equation is a pure quadratic.

A monic quadratic has the coefficient of \(x^2\) equal to 1. In option A, \(x^2+5x+6=0\) has coefficient 1 for \(x^2\), so it is monic. The closest distractor is option B because it is also a quadratic but has coefficient 2 for \(x^2\), so it is not monic. Exam tip: to identify a monic quadratic quickly, check only the coefficient of \(x^2\); if it is 1, the quadratic is monic.

Check by substitution: \(0^2+5\cdot0=0\), so \(x=0\) makes the equation zero and is therefore a root. Factoring gives \(x^2+5x=x(x+5)=0\), so the roots are \(x=0\) and \(x=-5\). Option C is wrong because \(x=5\) yields \(25+25\neq0\). Option D is misleading: \(x=-5\) is a root but not the only one. Exam tip: verify roots by direct substitution or factor the quadratic (look for a common factor first).

The discriminant is denoted by \(D\) and is given by \(D=b^2-4ac\). It determines the nature of the roots: two distinct real roots if \(D>0\), one repeated real root if \(D=0\), and two complex conjugate roots if \(D<0\). Option B has the wrong sign (\(+4ac\) instead of \(-4ac\)), while options A and D mix up the coefficients and are not the standard discriminant. Exam tip: correctly identify \(a,b,c\) from the equation and substitute into \(b^2-4ac\) to avoid sign errors.

When \(D=b^2-4ac=0\), the quadratic has two equal (repeated) real roots. The common root equals \(x=-\dfrac{b}{2a}\). Option A is a close distractor — two distinct real roots occur only when \(D>0\). Option C is wrong because no real roots occur only when \(D<0\). Option D is impossible since a quadratic cannot have three roots. Exam tip: compute \(D\) first and compare it with 0 to decide the nature of roots.

When (D<0), there are no real roots. A negative discriminant gives no roots on the real number line.

Answer: A, a = −1, b = 4, c = −1. Compare the equation with the standard quadratic form ax² + bx + c = 0. The coefficient attached to x² is a, the coefficient attached to x is b, and the term without x is c. In −x² + 4x − 1 = 0, the coefficient of x² is −1, so a = −1. The coefficient of x is +4, so b = 4. The constant term is −1, so c = −1. The sign before each term must be carried carefully. B changes the sign of a, even though the first term is negative. C changes the sign of b, although the x-term is positive. D rearranges the roles of the coefficients and constant instead of matching powers of x. Therefore A is the only complete and correct matching. Memory cue: read coefficients by descending powers: x² gives a, x gives b, and the number alone gives c.

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).