Roots of a Quadratic Equation

Correct answer: A. \(9\) and \(-9\)

Explanation: Rewriting \(x^2-81=0\) as \(x^2=81\) gives \(x=\pm 9\). Therefore, the two roots are \(9\) and \(-9\). Option B incorrectly treats 81 as a root, although 81 is the value of \(x^2\). In exams, remember to consider both the positive and negative values when taking the square root.

Correct answer: C. शून्य

Explanation: For \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Equal roots occur precisely when \(D=0\). If \(D>0\), there are two distinct real roots. Exam tip: identify the nature of roots by checking the discriminant first.

Correct answer: A. \(x^2-4x-12=0\)

Explanation: If the roots are 6 and −2, the equation is \((x-6)(x+2)=0\). Expanding it gives \(x^2-4x-12=0\), so option A is correct. Exam tip: the sum of the roots is 4, giving the coefficient −4 for \(x\), and their product is −12.

Correct answer: A. 13

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, \(a=1\) and \(b=-13\), so the sum is \(-\frac{-13}{1}=13\). The value 42 is related to the product of the roots, \(\frac{c}{a}\), not their sum. In an exam, use \(-b/a\) directly to find the sum.

Correct answer: A. \\(-2\\)

Explanation: For a quadratic equation \\(ax^2+bx+c=0\\), the product of its roots is \\(\\frac{c}{a}\\). Here, \\(a=7\\) and \\(c=-14\\), so the product is \\(\\frac{-14}{7}=-2\\). Option B has the sign wrong. Exam tip: remember that the sum and product of roots are \\( -\\frac{b}{a}\\) and \\(\\frac{c}{a}\\), respectively.

Correct answer: A. Constant term \(c\)

Explanation: For 0 to be a root, substituting \(x=0\) must satisfy the equation: \(5(0)^2+k(0)+c=0\). Hence, \(c=0\) is necessary. The coefficient \(k\) need not be zero. Exam tip: for a zero root, evaluate the polynomial at \(x=0\).

Correct answer: A. -3

Explanation: Substitute the given root \\(x=-3\\) into the equation: \\((-3)^2+u(-3)-18=0\\), which gives \\(9-3u-18=0\\). Thus, \\(-3u=9\\) and \\(u=-3\\). Therefore, option A is correct. Exam tip: When a root is given, substitute it directly into the polynomial and solve for the unknown parameter.

Correct answer: B. \((-8,-8)\)

Explanation: Since \(x^2+16x+64=(x+8)^2\), the equation becomes \((x+8)^2=0\). Hence \(x=-8\), so both roots are \((-8,-8)\). Option A results from an incorrect sign. Exam tip: when a quadratic is a perfect square, its repeated root can be found directly by setting the squared factor equal to zero.

Correct answer: A. 0, 5

Explanation: Factoring the equation gives \(4x^2-20x=4x(x-5)=0\). Thus, \(4x=0\) gives \(x=0\), and \(x-5=0\) gives \(x=5\). Therefore, the roots are 0 and 5. Exam tip: If the product of two factors is zero, set each factor equal to zero separately.

Correct answer: A. repeated root)

Explanation: Here, \(x^2-18x+81=(x-9)^2\). Thus, \((x-9)^2=0\) gives \(x=9\), and this root occurs twice. Therefore, the correct answer is 9. Exam tip: If a quadratic is a perfect square, set the linear expression inside the square equal to zero to find the repeated root.

Correct answer: C. \(x=0\) इसका एक मूल है।

Explanation: The equation \(x^2-100=0\) gives \(x^2=100=10^2\). Thus, \((x-10)(x+10)=0\), so its roots are \(10\) and \(-10\). Substituting \(x=0\) gives \(-100\neq0\), so statement C is incorrect. Exam tip: check a proposed root by substituting it directly into the equation.

Correct answer: C. \(4, -9\)

Explanation: By the zero-product property, either \((x-4)=0\) or \((x+9)=0\). Thus, \(x=4\) or \(x=-9\), so the roots are \(4\) and \(-9\). Option B has both signs wrong; in particular, \(x+9=0\) gives \(x=-9\). In an exam, set each factor equal to zero and solve separately.

Correct answer: B. −3

Explanation: The sum of the roots is found by adding them directly: 4 + (−7) = −3. Therefore, option B is correct. The value 3 represents only the difference in magnitudes, not the signed sum. Exam tip: do not confuse the sum of the roots with their product.

Correct answer: B. x² − 10x + 21 = 0

Explanation: For a monic quadratic with roots α and β, the equation is x² − (α + β)x + αβ = 0. This follows because (x − α)(x − β) expands to x² − (α + β)x + αβ. The given sum is 10 and product is 21, so substitution gives x² − 10x + 21 = 0. Therefore option B is correct. The minus sign before the x coefficient is essential: the coefficient is the negative of the root sum. Option A has the wrong sign, while options C and D place 21 in the x coefficient or use inappropriate signs. The equation also factors as (x − 3)(x − 7) = 0, confirming sum 10 and product 21.

Correct answer: B. \(4\)

Explanation: Substitute the given root \(x=4\) into the equation: \(2(4)^2-9(4)+n=0\). Thus, \(32-36+n=0\), which gives \(n=4\). Exam tip: When a root is given, substitute it directly into the quadratic equation to find the unknown constant.

Correct answer: A. 3 और 3

Explanation: Since \(x^2-6x+9=(x-3)^2\), we get \((x-3)^2=0\), so \(x=3\). Therefore, both roots are 3 and 3; the equation has equal roots. In an exam, identifying the perfect square form is the quickest method.

Correct answer: B. \(x-7\)

Explanation: By the factor theorem, if \(r\) is a root of a polynomial, then \((x-r)\) is its factor. Here, \(r=7\), so the required factor is \((x-7)\). The factor \((x+7)\) would correspond to the root \(-7\). Exam tip: substitute the given root directly into the form \(x-r\).

Correct answer: C. 0

Explanation: For every real \(x\), \(x^2\ge 0\), so \(x^2+36\ge 36\) and can never equal zero. Therefore, the equation has no real roots, making 0 the correct answer. As an exam tip, the discriminant can also be checked: \(D=b^2-4ac=-144<0\), which confirms that there are no real roots.

Correct answer: A. 3, −4

Explanation: Factoring the equation gives \(x^2+x-12=(x+4)(x-3)\). Thus, \(x+4=0\) gives \(x=-4\), and \(x-3=0\) gives \(x=3\). Therefore, the roots are 3 and −4. In option B, the signs of both roots are reversed. Exam tip: Look for two factors of −12 whose product is −12 and whose sum is 1.

Correct answer: A. \(x=11\) and \(x=-11\)

Explanation: The equation \(x^2=121\) means that the square of \(x\) is 121. Since \(11^2=121\) and \((-11)^2=121\), both \(x=11\) and \(x=-11\) are roots. Option B is incomplete because it includes only the positive root. Exam tip: When solving \(x^2=a\) for positive \(a\), consider both signs, \(x=\pm\sqrt{a}\).

Correct answer: B. 8

Explanation: Let the roots be 6 and \(x\). Since their product is 48, \(6x=48\), so \(x=\frac{48}{6}=8\). Therefore, the other root is 8. The value 42 results from subtracting 6 from 48, but the question gives the product, so division is required. Exam tip: Divide the product of the roots by the known root to find the other root.

Correct answer: B. 7

Explanation: The sum of the roots equals the first root plus the second root. Therefore, the other root = 12 − 5 = 7, so option B is correct. Exam tip: subtract the known root from the sum of the roots; do not add it.

Correct answer: C. \(r^2\)

Explanation: The two roots are \(r\) and \(r\), so their product is \(r \times r = r^2\). Option A, \(2r\), represents the sum of the roots, not their product. Exam tip: for equal roots \(r, r\), the sum is \(2r\) and the product is \(r^2\).

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

Explanation: The equation can be written as \(4x^2-12x+9=(2x-3)^2\). Hence, \((2x-3)^2=0\), so \(2x-3=0\) and \(x=\frac{3}{2}\). Both roots are equal and repeated. In the exam, identifying the perfect square is the quickest method; the negative value \(-\frac{3}{2}\) is not a root.

Correct answer: A. \(x^2-10x+25=0\)

Explanation: If both roots are 5, the equation formed from the roots is \((x-5)(x-5)=0\), or \((x-5)^2=0\). Expanding it gives \(x^2-10x+25=0\), so option A is correct. Option B has the wrong sign for the coefficient of \(x\), since the sum of the roots is 10. Exam tip: For roots \(\alpha\) and \(\beta\), a monic quadratic is \(x^2-(\alpha+\beta)x+\alpha\beta=0\).