Question

solve this by using standard 10 cbse syllabus

Answer 1

Quadratic equations ... Finding roots...

Q1.

  (x -1) (x-2) (3x-2) (3x+1) = 21
  (x² - 3x + 2) (9x² - 3x - 2) = 21
  9 x⁴ + x³ (-3*9 - 3*1) + x² (2*9 - 3*(-3) - 2*1) + x [-3*(-2) + 2(-3)] -2*2 = 21
  9 x⁴ - 30 x³ + 25 x² - 25 = 0
  
By looking at the first three terms, we can write now as:

   x² (9 x² - 30 x + 25) - 5² = 0
   x² (3 x - 5)² - 5² = 0        , factorize using  a² - b² = (a + b) (a - b)
   (3 x² - 5 x + 5) (3 x² - 5x - 5) = 0

So   3 x² - 5x + 5 = 0    givens 
       =>  x = [ 5 + √35 i ]/6       Two  roots imaginary.
 
    for 3x² - 5x - 5 = 0
        x = [5 + √85 ]/6     Real and negative roots.

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Q2.
   (x -1)(x+2)(x+3)(x+6) = 28
   (x² + x - 2) (x² + 9 x + 18) = 28
   x⁴ + x³ (1*9+1*1) + x² (1*18+1*9 -2*1) + x (-2*9+18*1) - 2*18 = 28
   x⁴ + 10 x³ + 25 x² - 64 = 0
   
  Looking at the first three terms we can write them as

    (x² + 5x)² - 8² = 0
=>  (x²+ 5x + 8) (x² + 5x - 8) = 0

For   x² + 5x + 8 = 0
           x = [-5 + √7 i ]/2      two complex roots.
  
For   x² + 5x - 8 = 0
          x = [ -5 + √57 ] /2      One positive and one negative root.

Answer 2

answer is upon our Level class 10 here
in this attachment