Roots of a Quadratic Equation

Correct answer: A. 0 is a repeated root

Explanation: The governing concept is multiplicity of roots and the discriminant. Because 5 is nonzero, divide 5x² = 0 by 5 to obtain x² = 0, or (x − 0)² = 0. Thus the only distinct root is x = 0, and it occurs twice in the factorisation; it is therefore a repeated, or equal, root. The discriminant confirms this: for 5x² + 0x + 0 = 0, D = b² − 4ac = 0² − 4(5)(0) = 0, which indicates two equal real roots. Option B confuses the coefficient with a root, option C gives values that do not satisfy the equation, and option D is false because zero is a real number. Hence option A is correct.

Correct answer: C. 2

Explanation: Factor the quadratic: \(x^2-2x = x(x-2)\). Setting each factor to zero gives the roots \(x=0\) and \(x=2\). Since only \(2\) appears among the options, it is the correct choice. Check a common distractor: for \(x=3\), \(3^2-2\cdot3=9-6=3\neq0\), so 3 is not a root. Exam tip: factorise first and then set each factor equal to zero to find roots quickly.

Correct answer: A. (x^2-3x-4=0)

Explanation: For a quadratic with roots α and β the monic form is x^2-(α+β)x+αβ=0. Here α=4 and β=−1 so α+β=3 and αβ=−4. Substituting gives x^2-3x-4=0, which factors as (x-4)(x+1)=0. The closest distractor C, x^2-5x+4=0, has roots 1 and 4 — it shares root 4 but not −1, so it is incorrect. Exam tip: compute sum and product of the given roots and form x^2-(sum)x+(product)=0.

Correct answer: C. −7

Explanation: The governing concept is Vieta’s sum-of-roots formula. For a quadratic equation Ax² + Bx + C = 0 with roots α and β, α + β = −B/A. In x² + 7x + 10 = 0, A = 1 and B = 7, so the sum is −7/1 = −7. Therefore option C is correct. A factorisation check gives x² + 7x + 10 = (x + 5)(x + 2), whose roots are −5 and −2; their sum is −7 as well. Option A ignores the negative sign in the formula. Option B is the constant term and is related to the product of the roots, not their sum. Option D is neither the correct sum nor the product. The coefficient comparison is sufficient and avoids unnecessary solving.

Correct answer: A. -1

Explanation: For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(\frac{c}{a}\). Here \(a=2\) and \(c=-2\), so product = \(\frac{-2}{2}=-1\). Option B (1) is wrong due to incorrect sign; option C (-2) wrongly uses \(c\) instead of \(c/a\). Exam tip: Always identify \(a\) and \(c\) first and be careful with signs.

Correct answer: A. \(c=0\)

Explanation: Substitute \(x=0\) into \(ax^2+bx+c=0\): this gives \(c=0\). Therefore \(x=0\) is a root only when the constant term \(c\) is zero. If \(c=0\), the equation reduces to \(ax^2+bx=0\) or \(x(ax+b)=0\), so \(x=0\) is indeed a root. Option B is incorrect because \(a=0\) makes the equation linear (not necessarily giving zero as a root); option C (\(b=0\)) does not ensure zero is a root unless \(c=0\) also; option D (\(a=b\)) is irrelevant to whether zero is a root. Exam tip: always test a suspected root by direct substitution into the equation.

Correct answer: B. (4)

Explanation: If a number is a root of a polynomial equation, substituting that number into the polynomial must make the result zero. The given root is -2, so replace every x in x^2 + kx + 4 = 0 by -2. Substitution must preserve the signs: (-2)^2 is positive 4, while k(-2) is -2k. This produces an equation involving only k.

The substitution gives 4 - 2k + 4 = 0, or 8 - 2k = 0. Hence -2k = -8 and k = 4. Therefore option B is correct. A common error is to treat (-2)^2 as -4, but squaring a negative number gives a positive result. Another error is to write +2k instead of -2k. The supplied answer correctly handles both signs.

Correct answer: B. (-3), (-3)

Explanation: Factor the quadratic: \(x^2+6x+9=(x+3)^2\). Hence \((x+3)^2=0\) gives the repeated root \(x=-3\). The discriminant \(\Delta=b^2-4ac=36-36=0\) confirms a double root. Option A is the sign-mistake counterpart; options C and D do not satisfy the sum and product required by the coefficients. Exam tip: check \(\Delta\) first—if it is zero, the equation has one repeated root.

Correct answer: A. (0) and (4)

Explanation: Factor the left side: \(3x^2-12x=3x(x-4)\). A product is zero when one factor is zero, so \(3x=0\) gives \(x=0\) and \(x-4=0\) gives \(x=4\). Hence the roots are 0 and 4. The closest distractor B (3 and 4) is wrong because substituting \(x=3\) yields \(3(3)^2-12(3)=27-36=-9\), not zero. Exam tip: always factor out the common factor first; if you divide by x, check x=0 separately.

Correct answer: B. (5) twice

Explanation: The polynomial factors as \(x^2-10x+25=(x-5)^2\), so the repeated root is \(5\). The closest distractor, \(-5\), would be correct only if the factor were \((x+5)^2\), which it is not. Quick exam tip: check for a perfect square or compute the discriminant \(b^2-4ac\); if it equals zero you have a repeated root.

Correct answer: C. 0

Explanation: Factor the equation: \((x-4)(x+4)=0\), so the roots are \(x=4\) and \(x=-4\). Substituting \(x=0\) gives \(0^2-16=-16\neq0\), so 0 is not a root. A common trap is \(\sqrt{16}\), which equals 4 and hence is a root; note that \(\sqrt{16}\) denotes the principal (positive) square root. Exam tip: always check by substitution or factoring to confirm whether a candidate value satisfies the equation.

Correct answer: C. 2 and -7

Explanation: Zero-product rule: if a product of two factors is zero, at least one factor is zero. So set \(x-2=0\) or \(x+7=0\). From \(x-2=0\) we get \(x=2\); from \(x+7=0\) we get \(x=-7\). Therefore the roots are 2 and −7. The closest distractor (A: 2 and 7) arises from taking the sign of the second factor incorrectly, so it is wrong. Exam tip: set each factor equal to zero separately and check signs of constants carefully.

Correct answer: B. -3

Explanation: Compute the sum directly: 1 + (−4) = −3. For a quadratic ax^2+bx+c=0 the sum of roots α+β = −b/a, which agrees with the direct addition here. The closest wrong option, −5, would arise if a student mistakenly treated 1 as −1 and computed (−1)+(−4)=−5. Exam tip: always write the signs explicitly and perform the addition carefully when a negative number is involved.

Correct answer: B. (x^2-6x+8=0)

Explanation: The standard form is (x^2-(\text{sum})x+\text{product}=0) so (x^2-6x+8=0). The sign of the sum term changes.

Correct answer: B. (1)

Explanation: If a number is a root of an equation, substituting that number into the equation must make the result zero. Here the given root is \(x=1\). Substitute it into \(2x^2-3x+m=0\): \(2(1)^2-3(1)+m=0\), so \(2-3+m=0\). This simplifies to \(m-1=0\), giving \(m=1\), which is option B.

The important point is that the root condition is checked by direct substitution. No quadratic formula or factorisation is needed. Testing the result, the equation becomes \(2x^2-3x+1=0\), and at \(x=1\) its value is \(2-3+1=0\). Thus the supplied answer is correct and the calculation is complete.

Correct answer: A. (1,2)

Explanation: Factorising gives \(x^2-3x+2=(x-1)(x-2)\), so the roots are \(x=1\) and \(x=2\). The closest distractor D contains 2 (which is correct) but also 3 (which is not), so D is wrong. Also the discriminant is \(b^2-4ac=9-8=1\), a positive perfect square, confirming two rational real roots. Exam tip: use Vieta — sum of roots =3 and product =2 to check your answer quickly.

Correct answer: B. \(x-2\)

Explanation: For any polynomial, a simple root \(r\) corresponds to the linear factor \(x-r\). So if the root is \(2\), the guaranteed factor is \(x-2\). Analysis of distractors: \(x+2\) would correspond to root \(-2\); \(2x+1\) corresponds to root \(-\tfrac{1}{2}\); and \(x^2+2\) is a quadratic expression and cannot be the linear factor associated with root \(2\). Exam tip: substitute \(x=2\) into the polynomial — it must give zero if 2 is truly a root.

Correct answer: C. 0

Explanation: Since \(x^2\ge 0\), we have \(x^2+4\ge 4>0\), so \(x^2+4\) can never be zero and there are no real roots. Using the quadratic discriminant: \(\Delta=b^2-4ac=0^2-4\cdot1\cdot4=-16<0\), which confirms no real solutions. Common distractor: students may think a quadratic always has two roots — that requires \(\Delta>0\). Exam tip: check \(\Delta\) or the minimum value of \(x^2\) to decide real roots quickly.

Correct answer: B. (2) and (-4)

Explanation: A quadratic equation can often be solved by factoring it into two linear factors. When a product is zero, at least one factor must be zero. This zero-product idea gives the possible values of the variable, called roots. The signs in the factors are important: a factor containing a plus sign gives a negative root, while a factor containing a minus sign gives a positive root.

Rewrite the expression as \\(x^2+2x-8=(x+4)(x-2)\\). Therefore, either \\(x+4=0\\), giving \\(x=-4\\), or \\(x-2=0\\), giving \\(x=2\\). Thus the roots are \\(-4\\) and \\(2\\), which matches option B. Substitution also confirms both values make the original equation equal to zero.

Correct answer: A. (4) and (-4)

Explanation: Taking square root of both sides of \(x^2=16\) gives \(x=\pm\sqrt{16}=\pm4\). So both +4 and -4 are roots. Option B is wrong because it omits the negative root (-4); C and D are incorrect since \(\sqrt{16}=4\), not 8, and 0 and 16 do not satisfy the equation. Exam tip: whenever you take square roots in equations, include the \(\pm\) sign to capture both roots.

Correct answer: B. 4

Explanation: Let the roots be α and β with product αβ = 12, and given α = 3. Thus β = \\(\frac{12}{3}=4\\). More generally for ax^2+bx+c=0 the product of roots equals c/a. Option A (3) is incorrect because 3×3 = 9 ≠ 12; options C and D (9 and 12) give products 27 and 36 respectively, so they also do not match. Exam tip: use product = (given product) ÷ (given root) to find the other root quickly and check by multiplication.

Correct answer: B. 3

Explanation: Let the roots be α and β with α+β = 5, and suppose one root α = 2. Then the other root is β = 5−2 = 3. Option A is wrong because if both roots were 2 the sum would be 4; options C and D do not satisfy the given sum 5. Exam tip: subtract the known root from the total sum and verify by adding the two roots.

Correct answer: B. (1)

Explanation: If the roots are (r) and (\frac{1}{r}) their product is (1). For reciprocal roots remember the product is (1).

Correct answer: A. (1) and (\frac{3}{2})

Explanation: (2x^2-5x+3=(2x-3)(x-1)) so the roots are (\frac{3}{2}) and (1). Be careful with factors that contain coefficients.

Correct answer: A. \(x^2+5x=0\)

Explanation: A is correct because \(x^2+5x = x(x+5)\), so \(x=0\) is a root. Rule: a quadratic has 0 as a root only if its constant term is 0 (i.e., the polynomial is divisible by x). For contrast, option C factors as \((x-2)(x-3)\) and D as \((x+1)^2\), so their roots are 2, 3 and −1 respectively, not 0. Exam tip: check the constant term quickly — if it is 0, then x=0 is a root.