Question

find k if x2 -2(1+3k)x+7(3+2k)=0has equal roots

Answer 1

★ QUADRATICS RESOLUTION ★

☣ FOR EQUAL ROOTS → D=0 OR b² = 4ac

☣ HENCE , IMPLYING THIS CRITERIA HERE ...

☣ GIVEN EQUATION → x² -2(1+3k)x + 7(3+2k) =0

☣ HENCE ...FOR EQUAL ROOTS ...

☣ 4 ( 1 + 3k )² = 4(7)(3+2k)

☣ FURTHERMORE REDUCTION YIELDS ...

☣ 9k² - 8k - 20 =0

☣ NOW , SOLVING THIS ACQUIRED QUADRATICS IN "k" YIELDS THE REQUIRED VALUES OF k

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★

Answer 2

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