Three consecutive positive integers k, l and m are such that l\(^2\) = 3(k+m). Find the value of m.
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7
Explanation
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Let k=x, L=x+1, M=x+2.Since L^2=3(K+M)
(X+1)^2=3(X+X+2)
Factorisn (x+1)^2=X^2+2X+1
So X^2+2X+1 =3(X+X+2)
X^2+2X+1=3(2X+2)
X^2+2X+1=6X+6
Collect lyk terms
X^2-4X-5=0
Solvin d quadratic equatn X=5=K
X+1=5+1=6=L
X+2=5+2=7=M
So M=7
Since no num ws given 4rm d equ. So Let k= x, l= x+1, m= x+2. 4rm d equ. l^2=3(k+m) , substutin d value of k,l and m in d equ. We av (x+1)^2=3(x+x+2)=use 2 to power on d left side nd use 3 to multiply on d right side , that is: x^2+1=3x+3x+6; x^2+1=6x+6 collect like terms x^2+6x=6-1; x^2+6x=5; x^2+6x-5=0 by factorization , we av x^2-x-5x-5=0:: x(x-1)-5(x+1) den we collect d outside one nd d one in bracket on d right side, we av (x-5)(x+1)=0 x-5=0, or x+1=0 wuch is x=5 or x=-1 therefore m=5-1=4 so d correct ans is 4 wuch is option A
m=7
Solution
If they are consecutive positive integers, then l=k+1 and m=k+2
Substitute (k+1) for l and (k+2) for m
in the equation l 2 =3(k+m)
(k+1) 2 =3(k+k+2)
k 2 +2k+1=6k+6
k 2 +2k-6k+1-6=0
k 2 -4k-5=0
Solve quadratically, k=5 or -1
The question says the integers are positive so k=5,l=5+1=6 and m=5+2=7
let k=x
l=x+1
m=x+2
from the equation
l^2=3(k+m)
(x+1)^2=3(x+x+2)
(x+1)(x+1)=3(2x+2)
x(x+1)+1(x+1)=6x+6
x^2 +x+x+1=6x+6
x^2 +2x+1=6x+6
make equation equal zero
x^2 +2x-6x+1-6=0
x^2 - 4x - 5=0
(x^2 + x)(-5x-5)=0
x(x+1)-5(x+1)=0
(x-5)(x+1)=0
x-5=0 or x+1=0
x=5 or x=-1
when x=-1
k=x=-1
l=x+1=-1+1=0
m=x+2=-1+2=1
when x=5
k=x=5
l=x+1=5+1=6
m=x+2=5+2=7
:- m=7
@myschool look into that solution
Let x be oranges
let y be mangoes
so,cost of oranges=5x
cost of mangoes=4y
since she bought as many mangoes as oranges,den 2x=y
therefore cost of mangoes=8x.
For atleast 65 and atmost 130
5x 8x=13x
65<_ 13x<_130
5<_x<_10
Sorry I think d ans is d since l=6.
Since dey are consecutive therefore l=k+m÷2.
K+m=2l
Therefore k+m=12
That's k=5 &m=7
The answer is 7.come to see it.
L square = 3[k plus m]
Let k = x , l= x plus 1 & m= x plus 2
[X plus 1] square = 3[x plus x plus 2]
X^2 plus 2x plus 1 = 6x plus 6
X^2 - 4x-5= 0
[X plus 1] (x-5)= 0
X= -1 or 5. So therefore since x = 5 =k
L= 6 m=7
k=x,l=x+1,m=x+2...substitute into d formula..l^2=3(k+m)....(x+1)^2=3(x+x+2)....(x+1)^2=x^2-4x-5= x=-1 or 5.....since m=x+2...m=-1+2=1....m=5+2=7....so d ans is 7-1=6..option c
3 constants;k:x,l:x+1,m:x+2 L*2=3(k+m),if u place d constants u'll get:(x+1)*2=3(x+x+2)=x*2+2x+1=6x+6,t will giv u x*2-4x-5=0,usin almighty formular u'll get;x=6 or x=10.usin d constant dat is m=x+2;m=6+2=8,10+2=12,therefore:12-8=4(A)
