Roots of a Quadratic Equation

The governing definition of a root, or zero, of a polynomial is a value of the variable that makes the polynomial equal to zero. Thus, if α is a root of p(x) = 0, substituting x = α into the polynomial gives p(α) = 0. The value p(α) cannot generally be declared positive or negative; that depends on the particular polynomial and the selected value. Option D is also incorrect because it says the entire polynomial p(x) equals α, whereas the definition requires evaluating p at α. The statement p(α)=0 is universally valid for every root. Hence option B is the only correct answer.

The governing operation is addition of the two given roots. The roots are −5 and 2, so their sum is (−5) + 2 = −3. Therefore option A is correct. The sign must be handled carefully: adding a positive number to a negative number means subtracting magnitudes, and the number with the larger absolute value determines the sign. Since |−5| is greater than |2|, the result is negative. Option B would result from reversing the sign or treating −5 as positive. Options C and D incorrectly add or subtract the magnitudes without applying the correct signed-number rule. The product of the roots would be −10, but that is not what the question asks.

The governing concept is the definition of a root or solution. A root of an equation is a value of the variable that makes the equation true after substitution. For a quadratic equation ax² + bx + c = 0, a number r is a root when ar² + br + c = 0. For example, in x² − 5x + 6 = 0, substituting x = 2 gives 4 − 10 + 6 = 0, so 2 is a root; x = 3 also works. A coefficient describes how a term is scaled, the constant term is the term independent of x, and the degree is the highest exponent. These identify features of the equation, not its root. Therefore option A is the only correct definition.

(p(2)=0) means the equation is satisfied when (x=2). Substitution is the key method in such questions.

Putting (x=2) gives (4-10+6=0) so it is a root. In exams always check the final sum after substitution.

The governing method is factorisation using the difference of two squares. Since 9 = 3², the equation becomes x² − 3² = 0. Applying u² − v² = (u − v)(u + v), we get (x − 3)(x + 3) = 0. By the zero-product property, at least one factor must be zero. Thus x − 3 = 0 gives x = 3, and x + 3 = 0 gives x = −3. Both values satisfy the original equation because their squares are 9. Therefore option A is correct. Option B mistakes the squared value for the roots, option C includes 0, for which the left side is −9, and option D omits the negative solution. A quadratic can have two distinct roots, so both signs must be considered.

(x^2-4x+3=(x-1)(x-3)) so the roots are (1) and (3). Factorisation is the fastest method for easy questions.

Factor the quadratic: \(x^2+5x+6=(x+2)(x+3)\). Setting each factor to zero gives \(x+2=0\Rightarrow x=-2\) and \(x+3=0\Rightarrow x=-3\). The closest distractor (B) simply has the signs reversed (+2 and +3), which is the common sign error. Exam tip: verify quickly using sum and product of roots — sum = \(-b/a\) = \(-5\) and product = \(c/a\) = \(6\).

With roots (0) and (5) the equation is (x(x-5)=0) that is (x^2-5x=0). Remember the form (x-r) while forming an equation from roots.

Core idea: substitute the given root directly into the equation. With \(x=1\) we get \(1 - k + 2 = 0\) or \(3 - k = 0\), hence \(k = 3\). Option B (2) often arises from a careless mistake such as equating the expression to the wrong value or an arithmetic slip when adding terms. Exam tip: always substitute the root and simplify step by step, watching signs carefully.

For a quadratic equation the sum of roots is (-\frac{b}{a}). Keep the sign of (b) correct while using the formula.

Divide the equation by \(a\) (assuming \(a\neq 0\)) to get \(x^2+\frac{b}{a}x+\frac{c}{a}=0\). Comparing with \(x^2-(\alpha+\beta)x+\alpha\beta=0\) gives \(\alpha\beta=\frac{c}{a}\). Option B has the wrong sign; options C and D confuse the coefficient \(b\) with the constant term. Exam tip: product of roots = constant term ÷ leading coefficient, and sum of roots = -(middle coefficient) ÷ leading coefficient.

The discriminant is defined by \(D = b^2 - 4ac\). Two equal (repeated) real roots occur exactly when \(D = 0\); the repeated root is \(-b/(2a)\). The closest distractor A (D > 0) is wrong because that case gives two distinct real roots; C (D < 0) gives complex conjugate roots; D (D = 1) is just a specific value and does not generally imply equal roots. Exam tip: always compute \(D=b^2-4ac\) and compare with zero to determine the nature of roots quickly.

Discriminant is defined by \(D=b^2-4ac\). Here \(a=1,\;b=-4,\;c=4\). Thus \(D=(-4)^2-4\times1\times4=16-16=0\). So the discriminant is 0 and the quadratic has equal (repeated) roots. Option A (4) is incorrect because correct substitution and arithmetic give 0, not 4. Exam tip: always list values of \(a,b,c\) explicitly before computing \(b^2-4ac\) to avoid sign or multiplication mistakes.

The discriminant \(D=b^2-4ac\) determines the nature of roots. If \(D>0\) there are two distinct real roots, if \(D=0\) there is one repeated real root; hence real roots occur exactly when \(D\ge 0\). Option A is incorrect because \(D<0\) gives complex (non-real) roots. Option C is just a specific value and not a general condition. Option D is incorrect because \(a=0\) makes the equation linear, not quadratic. Exam tip: compute \(D=b^2-4ac\) first to decide root nature quickly.

(x^2) is never negative so (x^2+1=0) is not satisfied by any real number. Sign checking helps solve such questions quickly.

Substituting \(x=0\) gives \(3(0)^2+2(0)=0\). Hence the left-hand side is zero and \(x=0\) is a root. Alternatively factor: \(3x^2+2x=x(3x+2)=0\), giving roots \(x=0\) and \(x=-\tfrac{2}{3}\). Option (C) is incorrect because substituting \(x=2\) yields \(3(2)^2+2(2)=16\neq0\). Option (D) is wrong since the equation is explicit and substitution determines whether a value is a root. Exam tip: To check a root substitute the value into the polynomial or factor the polynomial to find all roots quickly.

Write the equation as \(x^2-x=0\) and factorize: \(x(x-1)=0\). Hence the roots are \(x=0\) and \(x=1\). Option B (1 and 2) is wrong because substituting \(x=2\) gives \(4\neq2\). Option C (-1 and 1) is wrong since \(x=-1\) does not satisfy the equation (\((-1)^2=1\neq-1\)). Exam tip: always bring the equation to \(=0\) and factorize to find roots quickly and reliably.

Divide both sides by 2: from \(2x^2-8=0\) we get \(x^2=4\). Taking square roots gives \(x=\pm 2\), i.e. 2 and −2. Option B would come from incorrectly assuming \(x^2=16\); option C is wrong because 0 does not satisfy the equation; option D is incomplete because it omits the negative root. Exam tip: when taking square roots always include both \(+\) and \(-\) solutions after squaring.

Factor the quadratic: \(x^2+2x+1=(x+1)^2\). Thus \((x+1)^2=0\) gives the double root \(x=-1\). A common mistake is choosing \(1\), which would correspond to \((x-1)^2\) and the polynomial \(x^2-2x+1\). Exam tip: check either factorisation or compute the discriminant \(b^2-4ac\); if it equals 0, the quadratic has a repeated root.

The governing construction is the equation formed from known roots. If α and β are the roots of a monic quadratic, its equation is (x − α)(x − β) = 0. Substituting α = 2 and β = 5 gives (x − 2)(x − 5) = 0. On expansion, x² − 5x − 2x + 10 = x² − 7x + 10, so the required equation is x² − 7x + 10 = 0, which is option A. The same result follows from Vieta’s relations: the sum of roots is 7, so the coefficient of x is −7, and the product is 10, so the constant term is 10. Option B has the wrong sign for the sum, option C has the wrong sum, and option D has incorrect signs and product.

Use factorisation or sum-and-product of roots. Factorising gives \(x^2-6x+8=(x-4)(x-2)\), so with one root 4 the other root is 2. The close distractor 1 is wrong because \(4\times1=4\), not equal to the constant term 8. Exam tip: For \(ax^2+bx+c=0\), use \(\alpha+\beta=-\dfrac{b}{a}\) and \(\alpha\beta=\dfrac{c}{a}\) to find the other root quickly.

A negative product occurs when one root is positive and the other is negative. (\alpha\beta<0) is a quick sign check.

By the factor theorem, if \(x-a\) is a factor then substituting \(x=a\) into the polynomial gives value 0. Hence \(a\) is a root (zero) of the polynomial. Option A (coefficient) is incorrect because a coefficient is the multiplier of a term and does not indicate a value that makes the polynomial zero. Option D (constant term) is a different concept — the term independent of \(x\). Exam tip: plug in \(x=a\); if the polynomial evaluates to 0, then \(x-a\) is a factor.

Substitute the root into the equation: with \(x=3\) we get \(3^2+3p-6=0\), so \(9+3p-6=0\) ⇒ \(3+3p=0\) ⇒ \(p=-1\). A common wrong choice (1) comes from an arithmetic or sign mistake when simplifying; check signs carefully. Exam tip: always substitute the given root directly and simplify step by step to avoid sign errors.