Roots of a Quadratic Equation

Correct answer: A. 5

Explanation: When the discriminant \(D=0\), the two roots are equal. Let each root be \(x\). Then \(x+x=10\), so \(2x=10\) and \(x=5\). Therefore, each root is 5. Exam tip: When \(D=0\), divide the sum of the roots by 2 to find the repeated root.

Correct answer: A. 6

Explanation: Factoring gives \(x^2-4x-5=(x-5)(x+1)\), so the roots are \(5\) and \(-1\). Therefore, their absolute difference is \(|5-(-1)|=6\). Exam tip: when subtracting a negative root, carefully account for the two minus signs; the coefficient \(4\) is not the difference between the roots.

Correct answer: A. \(3\)

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=2\) and \(b=-6\), so \(\alpha+\beta=-\frac{-6}{2}=3\). Option B results from mishandling the negative sign of \(b\). Exam tip: identify the coefficients carefully before applying \(-b/a\).

Correct answer: B. दो वास्तविक और समान मूल

Explanation: Here \(a=1, b=-6, c=9\). The discriminant is \(D=b^2-4ac=(-6)^2-4(1)(9)=0\), so the roots are real and equal. Distinct real roots occur only when \(D>0\). Exam tip: check the discriminant first to identify the nature of roots.

Correct answer: A. \(k\le 1\)

Explanation: A quadratic equation has real roots when its discriminant satisfies \(D=b^2-4ac\ge0\). Here, \(a=1\), \(b=2\), and \(c=k\), so \(D=2^2-4(1)(k)=4-4k\). Therefore, \(4-4k\ge0\), which gives \(k\le1\). Hence, option A is correct. If \(k>1\), as stated in option B, the discriminant becomes negative, so the roots are not real. Exam tip: For real roots of a quadratic, always begin with the condition \(D\ge0\).

Correct answer: A. Both negative

Explanation: The roots are (-5) and (-2) because ((x+5)(x+2)=0). When the sum is negative and product is positive, both roots are negative.

Correct answer: A. \(-6\)

Explanation: For the quadratic equation \(x^2+ax+9=0\), the sum of the roots is \(-\frac{a}{1}=-a\). Since the roots are \(3\) and \(3\), their sum is \(6\). Thus, \(-a=6\), giving \(a=-6\). The value \(6\) results from missing the negative sign. In exams, remember that the sum of roots of \(x^2+bx+c=0\) is \(-b\).

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

Explanation: Factorise the quadratic: \(6x^2-7x+2=(3x-2)(2x-1)\). Thus, \((3x-2)(2x-1)=0\) gives \(x=\frac{2}{3}\) or \(x=\frac{1}{2}\). Therefore, option A is correct. The values in option D do not satisfy the two linear factors. In an exam, set each linear factor equal to zero after factorisation to obtain the roots.

Correct answer: A. (-a^2)

Explanation: For a quadratic equation (a_1x^2+bx+c=0), the product of its roots equals (\frac{c}{a_1}). Here, the product is (a\times(-a)=-a^2), so (\frac{c}{a_1}=-a^2). Therefore, option A is correct. Exam tip: When the roots have equal magnitudes and opposite signs, their product is the negative of the square of either root.

Correct answer: A. (x^2+3x-10=0)

Explanation: For roots (alpha) and (beta), the monic quadratic equation is (x^2-(alpha+beta)x+alpha beta=0). Substituting the given values gives (x^2-(-3)x-10=0), which simplifies to (x^2+3x-10=0). Option B has the wrong sign for the sum of the roots. Exam tip: use the pattern (x^2-(sum of roots)x+product of roots=0).

Correct answer: A. 26

Explanation: Let the roots be \(\alpha\) and \(\beta\). By Vieta’s formulas, \(\alpha+\beta=6\) and \(\alpha\beta=5\). Therefore, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=6^2-2(5)=36-10=26\). Hence, option A is correct. Remember that 36 is the square of the sum of the roots, not the sum of their squares.

Correct answer: A. (2)

Explanation: Putting (x=2) gives (4a-10+2=0), so (4a=8) and (a=2). Substitute in questions with an unknown coefficient.

Correct answer: A. 14 and 49

Explanation: Here, \(a=1\), \(b=-14\), and \(c=49\). For a quadratic equation, the sum of the roots is \(-b/a=14\), while their product is \(c/a=49\). Indeed, \(x^2-14x+49=(x-7)^2\), so both roots are 7; their sum is 14 and their product is 49. Exam tip: directly apply the relations sum \(=-b/a\) and product \(=c/a\). Option C gives the individual repeated root instead of its sum and product.

Correct answer: A. 0

Explanation: For a quadratic ax^2+bx+c=0 the discriminant is \(D=b^2-4ac\). If \(D<0\), there are no real roots; instead the equation has two complex conjugate roots. The closest distractor C (2) would be correct only when \(D>0\), which gives two distinct real roots; when \(D=0\) there is one repeated real root. Exam tip: compute the discriminant first — its sign immediately tells you the number of real roots (D>0 → 2, D=0 → 1, D<0 → 0).

Correct answer: A. (-1)

Explanation: Here (\alpha+\beta=\frac{4}{3}) and (\alpha\beta=-\frac{4}{3}). Therefore (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-1).

Correct answer: A. Root

Explanation: Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.

Correct answer: A. Yes

Explanation: To check whether a number is a root, substitute \(x=4\) into the equation: \(4^2-9(4)+20=16-36+20=0\). Since the value of the polynomial is zero, \(x=4\) is a root. In fact, \(x^2-9x+20=(x-4)(x-5)\), so the two roots are \(4\) and \(5\); therefore, the statement that only \(x=5\) is a root is incorrect. Exam tip: a number is a root of a polynomial equation exactly when substitution gives zero.

Correct answer: A. Yes

Explanation: To check whether a number is a root, substitute it into the equation and verify whether the left-hand side becomes zero. For x = -2, 3(-2)² + 2(-2) - 8 = 12 - 4 - 8 = 0. Therefore, x = -2 is a root of the equation. Option C is incorrect because substituting x = 2 gives 12 + 4 - 8 = 8, not zero. Exam tip: the square of a negative number is positive, but its linear term remains negative.

Correct answer: A. (5, 7)

Explanation: Factorise the quadratic: \(x^2-12x+35=(x-5)(x-7)\). Thus, \((x-5)(x-7)=0\) gives \(x=5\) or \(x=7\), so the roots are \((5,7)\). Option B has numbers whose product is 35, but their sum is not 12. Exam tip: For \(x^2+bx+c\), look for two numbers whose sum is \(-b\) and product is \(c\).

Correct answer: A. 4 and \(\frac{1}{3}\)

Explanation: The quadratic factors as \(3x^2-13x+4=(3x-1)(x-4)\). Thus, \((3x-1)(x-4)=0\) gives \(x=\frac{1}{3}\) or \(x=4\), so option A is correct. The values in option D have sum \(\frac{7}{3}\), whereas the sum of the roots must be \(\frac{13}{3}\). Exam tip: verify the roots using the sum \(-\frac{b}{a}\) and product \(\frac{c}{a}\).

Correct answer: A. \(-4,\ -\frac{1}{2}\)

Explanation: Factoring the quadratic gives \(2x^2+9x+4=(2x+1)(x+4)\). Thus, \((2x+1)(x+4)=0\) gives \(x=-\frac{1}{2}\) or \(x=-4\), so option A is correct. In option B, the signs of both roots are incorrect. Exam tip: Set each linear factor equal to zero to obtain the roots directly.

Correct answer: A. x² − 3x − 18 = 0

Explanation: If α and β are the roots of a quadratic equation, its equation is (x − α)(x − β) = 0. Here α = −3 and β = 6, so (x + 3)(x − 6) = 0. Expanding gives x² − 3x − 18 = 0, making option A correct. Exam tip: the sum of the roots is 3, so the coefficient of x is −3, while their product is −18, the constant term.

Correct answer: A. 3

Explanation: Since \(x=3\) is a root, substitute it into the equation: \(3^2+3k-18=0\). Thus, \(9+3k-18=0\), so \(3k=9\) and \(k=3\). Hence, option A is correct. Option B results from an incorrect sign. Exam tip: substitute the given root directly into the quadratic equation and solve for the parameter.

Correct answer: A. \(\frac{11}{4}\)

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, \(a=4\) and \(b=-11\), so the sum is \(-\frac{-11}{4}=\frac{11}{4}\). Option B has the wrong sign, while the options containing \(\frac{7}{4}\) reflect confusion with the constant term. Exam tip: use \(-b/a\) directly to find the sum of the roots.

Correct answer: A. \\(-\\frac{8}{5}\\)

Explanation: For a quadratic equation \\(ax^2+bx+c=0\\), the product of its roots is \\(\\frac{c}{a}\\). Here, \\(a=5\\) and \\(c=-8\\), so the product is \\(\\frac{-8}{5}=-\\frac{8}{5}\\). Option B has the wrong sign. Exam tip: remember that the sum of the roots is \\( -\\frac{b}{a} \\), while their product is \\( \\frac{c}{a} \\).