#### Ram is thrice as old as Shyam. The product of their ages 2 years from now will be 84. The present age of Ram is ____.

Let the present age of Shyam be x years. Therefore, present age of Ram = 3 x years their ages two years from now will be (x + 2) years and (3x + 2) years respectively. Product of their ages two years from now = 84 ⇒ (x + 2)(3x + 2) = 84 ⇒ 3 x^{2} + 2x + 6x + 4 = 84 ⇒ 3 x^{2}– 12x + 20x – 80 = 0 ⇒ 3x(x – 4) + 20(x – 4) = 0 ⇒ (x – 4)(3x + 20) = 0 ⇒ x = 4 years (∵ x can’t be negative) Present age of Ram = 3x = 3 × 4 = 12 years

#### If (k^{2} – 1) x^{2} + 3kx + 2 = 0 has real and equal roots, then k is a real number.

Given, (k^{2} – 1) x^{2} + 3kx + 2 = 0 ----- (1) Here, a = k^{2} – 1, b = 3k, c = 2 Also, given that The equation has real and equal roots ⇒ b^{2} –4ac = 0 ⇒ 9 k^{2} – 4(k^{2} – 1)2 = 0 ⇒ 9 k^{2} – 8 k^{2} + 8 = 0 ⇒ k^{2} + 8 = 0 ⇒ k^{2} = -8 Hence, k is not a real number.

#### Roots of the equation x^{2} – 4x – 21 = 0 are ____.

Given, x^{2} – 4x – 21 = 0 ⇒ x^{2} – 7x + 3x – 21 = 0 ⇒ x(x – 7) + 3(x – 7) = 0 ⇒ (x – 7)(x + 3) = 0 ⇒ x = 7 or x = -3 Hence, the roots are 7, −3

#### The product of Joseph’s age three years ago and his age 2 years hence is 6 more than three times his present age. The present age of Joseph is ____.

Let the present age of Joseph be x years Age of Joseph three years ago = (x − 3) years Age of Joseph two years after = (x + 2) years As per the problem, (x – 3)(x + 2) = (3x + 6) ⇒ x^{2} + 2x – 3x – 6 = 3x + 6 ⇒ x^{2}– x – 6 -3x – 6 = 0 ⇒ x^{2} – 4x – 12 = 0 ⇒ x^{2}– 4x -12 = 0 ⇒ x^{2} – 6x + 2x – 12 = 0 ⇒x(x -6) + 2(x – 6) = 0 ⇒(x – 6)(x + 2) ⇒ x = 6 or x = -2 ⇒ x = 6 years (∵ Age of a person cannot be negative) Hence, the present age of Joseph is 6 years.

#### Roots of the equation -2 x^{2} + 4x – 10 = 0 are real numbers.

Given, -2 x^{2} + 4x – 10 = 0 Here, a = -2, b = 4, c = -10 b^{2} – 4ac = 16 -4(-2)(-10) = -64 < 0 Hence, no real roots exist for the given quadratic equation.

#### The quadratic equations for (x^{2} + 2)^{2} = x^{4} + 2x is ____.

Given, (x^{2 }+ 2)^{2} = x^{4} + 2x ⇒ x^{4} + 4 + 4x^{2} = x^{4} + 2x ⇒ 4x^{2} – 2x + 4 = 0 Dividing throughout by 2, we get 2x^{2} – x + 2 = 0

#### If the product of two consecutive odd negative integers is 143, then their sum is ____.

Let the smaller of the odd negative integers be x Then other integer is (x + 2) Given that their product is 143. ⇒ x(x + 2) = 143 ⇒ x^{2} + 2x – 143 = 0 ⇒ x^{2} + 13x – 11x – 143 = 0 ⇒ x(x + 13) – 11(x + 13) = 0 ⇒ x(x + 13)(x – 11) = 0 ⇒ x = -13, 11 ⇒ x = -13 (∵ x is a negative integer) Then, the other number = −13 + 2 = -11 Hence, the required sum is −24.

#### ‘A’ has 2 balls more than ‘B’ has. If the product of the number of balls with them is 8, then the number of balls with ‘A’ is ____.

Let the number of balls with B be x. ⇒ Number of balls with A =(x+2) Given that the product of the number of balls with A and B is 8. ⇒ x(x + 2) = 8 ⇒ x^{2} + 2x – 8 = 0 ⇒ x^{2} + 4x – 2x – 8 = 0 ⇒ x(x – 4) – 2(x + 4) = 0 ⇒ (x + 4)(x – 2) = 0 ⇒ x =2, x = -4 since x cannot be negative ⇒ x = 2 ⇒ x + 2 = 4 Hence, the number of balls with A is 4.

#### If the sum of the squares of two consecutive natural numbers is 41, then the numbers are ____.

Let the smaller number be x, Then the other number is (x + 1) (∵ the numbers are consecutive) ⇒ x^{2} + (x + 1)^{2} = 41 ⇒ x^{2} + x^{2 }+2x + 1= 41 ⇒ 2 x^{2} +2x – 40 = 0 ⇒ x^{2} + x – 20 = 0 ⇒ x^{2} + 5x – 4x – 20 = 0 ⇒ x(x + 5) – 4 (x + 5) = 0 ⇒ (x + 5)(x – 4) = 0 ⇒ x = -5 or x = 4 ⇒ x = 4 (∵ x can’t be negative) ⇒ x + 1 = 5 Hence, the required numbers are 4 and 5.

#### The quadratic equation for x^{2} /2 +(x + 2)^{2}= 5 is ____.

Given, x^{2} /2 + (x + 2)^{2 }= 5 ⇒ x^{2} /2 + x^{2} + 4x +4 = 0 ⇒ x^{2}+ 2x^{2}+ 8x +8 = 10 ⇒ 3 x^{2}+ 8x – 2 = 0