Quantitative Aptitude — Quadratic Equations

The given quadratic equation is x^2 - 5x + 6 = 0. We can factorize this equation: x^2 - 2x - 3x + 6 = 0 x(x - 2) - 3(x - 2) = 0 (x - 2)(x - 3) = 0 Setting each factor to zero gives the roots: x - 2 = 0 => x = 2 x - 3 = 0 => x = 3 Thus, the roots are 2 and 3.

The given quadratic equation is 2x^2 - 7x + 3 = 0. Using the quadratic formula, x = [-b +/- sqrt(b^2 - 4ac)] / 2a. Here, a = 2, b = -7, c = 3. Discriminant D = b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25. x = [ -(-7) +/- sqrt(25) ] / (2 * 2) x = [ 7 +/- 5 ] / 4 Two roots are: x1 = (7 + 5) / 4 = 12 / 4 = 3 x2 = (7 - 5) / 4 = 2 / 4 = 1/2 Thus, the roots are 3 and 1/2.

For a quadratic equation ax^2 + bx + c = 0 to have equal roots, its discriminant (D) must be zero. D = b^2 - 4ac = 0 Given equation: x^2 - 4x + k = 0. Here, a = 1, b = -4, c = k. Substitute these values into the discriminant formula: (-4)^2 - 4(1)(k) = 0 16 - 4k = 0 4k = 16 k = 16 / 4 k = 4 Thus, for k = 4, the equation will have equal roots.

For a quadratic equation ax^2 + bx + c = 0, the sum of roots (alpha + beta) = -b/a and the product of roots (alpha * beta) = c/a. Given equation: 3x^2 - 5x + 2 = 0. Here, a = 3, b = -5, c = 2. Sum of roots (alpha + beta) = -(-5)/3 = 5/3. Product of roots (alpha * beta) = 2/3. We need to find (alpha + beta) + (alpha * beta): (5/3) + (2/3) = (5 + 2) / 3 = 7/3. Thus, the value is 7/3.

If alpha and beta are the roots of a quadratic equation, the equation can be written as x^2 - (alpha + beta)x + (alpha * beta) = 0. Given roots are 3 and -5. Sum of roots (alpha + beta) = 3 + (-5) = -2. Product of roots (alpha * beta) = 3 * (-5) = -15. Substitute these values into the formula: x^2 - (-2)x + (-15) = 0 x^2 + 2x - 15 = 0 Thus, the quadratic equation is x^2 + 2x - 15 = 0.

For a quadratic equation ax^2 + bx + c = 0 to have real roots, its discriminant (D) must be greater than or equal to zero. D = b^2 - 4ac >= 0 Given equation: x^2 + mx + 9 = 0. Here, a = 1, b = m, c = 9. Substitute these values into the discriminant inequality: m^2 - 4(1)(9) >= 0 m^2 - 36 >= 0 m^2 >= 36 Taking the square root of both sides: sqrt(m^2) >= sqrt(36) |m| >= 6 This implies that m must be greater than or equal to 6, or m must be less than or equal to -6. So, m >= 6 or m

Let the roots of the quadratic equation x^2 - 8x + k = 0 be alpha and beta. According to the problem, one root is three times the other. So, let beta = 3 * alpha. For a quadratic equation ax^2 + bx + c = 0: Sum of roots (alpha + beta) = -b/a Product of roots (alpha * beta) = c/a From the given equation, a = 1, b = -8, c = k. Sum of roots: alpha + 3*alpha = -(-8)/1 4*alpha = 8 alpha = 8 / 4 alpha = 2 Now, find the second root: beta = 3 * alpha = 3 * 2 = 6. The roots are 2 and 6. Product of roots: alpha * beta = k/1 2 * 6 = k k = 12 Thus, the value of k is 12.

Let the roots of the quadratic equation ax^2 + bx + c = 0 be alpha and beta. According to the problem, the roots are reciprocal to each other. This means if one root is alpha, the other root is 1/alpha. For a quadratic equation, the product of roots (alpha * beta) = c/a. Substitute the reciprocal roots into the product formula: alpha * (1/alpha) = c/a 1 = c/a Multiplying both sides by 'a': a = c Thus, if the roots are reciprocal, then a = c.

For the first equation: x^2 + px + q = 0 Roots are alpha and beta. Sum of roots: alpha + beta = -p/1 = -p (Formula: -b/a) Product of roots: alpha * beta = q/1 = q (Formula: c/a) For the second equation: x^2 + rx + s = 0 Roots are (alpha + beta) and (1/alpha + 1/beta). Let the first root be X1 = alpha + beta. Let the second root be X2 = 1/alpha + 1/beta. From the first equation, we know X1 = -p. Now, calculate X2 using the roots of the first equation: X2 = (alpha + beta) / (alpha * beta) X2 = (-p) / q For the second equation, the product of roots is s/1 = s. So, s = X1 * X2 s = (-p) * (-p/q) s = p^2 / q Thus, the value of s is p^2/q.

Let the number be x. Its reciprocal is 1/x. According to the problem, the sum of the number and its reciprocal is 13/6: x + 1/x = 13/6 To solve this, multiply the entire equation by 6x to clear the denominators: 6x * (x + 1/x) = 6x * (13/6) 6x^2 + 6 = 13x Rearrange the terms to form a standard quadratic equation (ax^2 + bx + c = 0): 6x^2 - 13x + 6 = 0 Now, we can solve this quadratic equation using factorization or the quadratic formula. Using factorization: We need two numbers whose product is 6*6 = 36 and sum is -13. These numbers are -9 and -4. 6x^2 - 9x - 4x + 6 = 0 3x(2x - 3) - 2(2x - 3) = 0 (3x - 2)(2x - 3) = 0 Setting each factor to zero: 3x - 2 = 0 => 3x = 2 => x = 2/3 2x - 3 = 0 => 2x = 3 => x = 3/2 Thus, the number is 2/3 or 3/2.