English

Solve the following quadratic equation: x^2 – (2b – 1)x + (b^2 – b – 20) = 0

Advertisements
Advertisements

Question

Solve the following quadratic equation:

x2 – (2b – 1)x + (b2 – b – 20) = 0 

Sum
Advertisements

Solution

We write, –(2b – 1)x = –(b – 5)x – (b + 4)x as 

x2 × (b2 – b – 20) = (b2 – b – 20)x2 = [–(b – 5)x] × [–(b + 4)x] 

∴ x2 – (2b – 1)x + (b2 – b – 20) = 0 

⇒ x2 – (b – 5)x – (b + 4)x + (b + 4) = 0 

⇒ x[x – (b – 5)] – (b + 4)[x – (b – 5)] = 0 

⇒ [x – b – 5][x – (b + 4)] = 0 

⇒ x – (b – 5) = 0 or x – (b + 4) = 0 

⇒ x = b – 5 or x = b + 4 

Hence, b – 5 and b + 4 are the roots of the given equation.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Quadratic Equations - EXERCISE 4A [Page 183]

APPEARS IN

R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Quadratic Equations
EXERCISE 4A | Q 44. | Page 183
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×