Math, asked by Ramu111, 1 year ago

if x^2-y^2+4x-6y+k is resolvable into two linear factors , then k=

Answers

Answered by mysticd
15
x² -y²+4x-6y+k
= x² +4x -y² -6y +k
= x² + 4x +2² - (y² +6y +3²) ------(1)
= (x+2)² - (y+3)²
= [x+2+y+3] [ x+2 -y-3]
=(x+y+5) ((x-y-1)

from (1) 
k = 2² -3² = 4-9 =-5

mysticd: u'r welcome
Answered by abhi178
18
x² - y² +4x -6y + K is resolvable into two linear factor ,
this is possible only when ,
∆ = 0
where ax² +by² +2hxy + 2gx +2fy + c = 0 is an standard form of above equation .
and ∆ = abc +2fgh -af² -bg² -ch² =0
then this is resolvae into two linear equations or factors .

now,
convert the equation of an standard form 1.x² + (-1).y² + 2.0.xy + 2.2.x + 2.(-3).y + K
now,

∆ = 1 ×- 1 × K +2× (-3)×2×0- 1.(-3)² +1.(2)² -K(O)² =0

-K + 0 -9 +4 = 0

-K -5 = 0

K = -5
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