Question

Find the value of ‘k’ for which the given equation has real roots: x2 – 2x ( 1 + 3k) + 7(3+2k) = 0

Answer 1

Answer: -10/9 and 2

Step-by-step explanation: hello users ......

we have given equation is  

x² + 2x(1 + 3k) +7(3 + 2k) = 0 

or 

x² + 2(1 + 3k)x +7(3 + 2k) = 0 

roots are equal 

we have to find the value of k =? 

now 

we know that an equation have equal roots if and only if D = 0

=> b² -4ac = 0 

here comparing the equation with 

ax² + bx + c = 0 

here 

a= 1, b =  2(1 + 3k) and c=  7(3 + 2k) 

now 

   b² - 4ac = 0

 

=  [ 2(1 + 3k) ]² - 4 ×1×7(3 + 2k)  = 0 

= 4×(1 + 3k)² - 4×7(3 + 2k) = 0

 = 4(1 + 9k² + 6k) - 4(21 +14k) =0

= (1 + 9k² + 6k) - (21 +14k) =0

=> 9k² - 8k -20 = 0 

   now solving the equation we get 

= 9k² - 18k + 10k - 20 = 0

= 9k( k- 2)  + 10(k -2) =0 

=> 9k +10=0 and k -2 = 0

=> k = -10/9 and 2 answer 

☆☆ hope it helps ☆☆