Question

Find k so that one root of the equation 2kx2–20x + 21 = 0 exceed the other by 2.
step by step explanation please

Answer 1

Answer:

Step-by-step explanation:

let the two roots be x and x+2

2x+2=20/2k=10/k

2x=(10/k)-2

2x=(10-2k)/k

x=(5-k)/k

(x)(x+2)=21/2k

((5-k)/k)((5-k)/k+2)=21/2k

((5-k)/k)(5-k+2k)/k=21/2k

(5-k)(5+k)/k^2=21/2k

25-k^2=21k/2

50-2k^2=21k

2k^2+21k-50=0

2k^2-4k+25k-50=0

2k(k-2)+25(k-2)=0

2k=-25,k=2

k=-25/2

Answer 2

Step-by-step explanation:

2x

2

+kx−2=0

1) if one root os -2 then

if satisfy the equation

2(−2)

2

+k(−2)−2=0

8−2−2k=0

k=

2

6

=3