Question

If x=-4 is a root of the equation
x^2+2x+4p=0, find the values
of k for which the equation
x^2+px(1+3K)+7(3+2k)=0 has
equal roots.

Answer 1

put x= -4 in 1st equation
16-8+4p=0
4p=-8
p= -2
put p= -2 in 2nd equation
${x}^{2} + ( - 4)x(1 + 3k) + 7(3 + 2k) = 0 \\ {x}^{2} - 4(1 + 3k)x + 7(3 + 2k) = 0 \\ since \: roots \: are \: equal \\ so \: \: \: \: d = 0 \\ {b}^{2} - 4ac = 0 \\ { (- 4(1 + 3k))}^{2} - 4 \times 1 \times 7(3 + 2k) = 0 \\ 16(1 + 9 {k}^{2} + 6k) - 84 - 56k = 0 \\ 144 {k}^{2} + 40k - 68 = 0 \\ 4(36 {k}^{2} + 10k - 17) = 0$
now use quadratic formula and find the value of k

Answer 2

Answer:

put x= -4 in 1st equation

16-8+4p=0

4p=-8

p= -2

put p= -2 in 2nd equation  

{x}^{2}  + ( - 2)x(1 + 3k) + 7(3 + 2k) = 0

{x}^{2}  - 2(1 + 3k)x + 7(3 + 2k) = 0

since  roots  are  equal,  so  d = 0  i.e.,  {b}^{2}  - 4ac = 0

{ (-2(1 + 3k))}^{2}  - 4 x 1 x 7(3 + 2k) = 0

4(1 + 9 {k}^{2}  + 6k) - 84 - 56k = 0

36 {k}^{2}  - 32k - 80 = 0

9k^2-8k-20=0

k=2 or k=-10/9