Solve for x. 9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0
Answers
9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0
9x^2 - 9(a + b)x + [2a(a + 2b) + b(a + 2b)] = 0
9x^2 - 9(a + b)x + [(a + 2b)(2a + b)] = 0
9x^2 – 3[(a + 2b) + (2a + b)]x + [(a + 2b)(2a + b)] = 0
9x^2 – 3(a + 2b)x - 3(2a + b)x + [(a + 2b)(2a + b)] = 0
3x[3x – (a + 2b)] - (2a + b)[3x + (a – 2b)] = 0
[3x – (a + 2b)][3x - (2a + b)] = 0
[3x – (a + 2b)] = 0
or [3x - (2a + b)] = 0
3x = (a + 2b) or
3x = (2a + b)
x = (a + 2b)/3 or
x = (2a + b)/3
Step-by-step explanation:
Let P,Q,R and S are the midpoint of AB,BC,CD andDA.
∴ Co-ordinates of P=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
−1−1
,
2
−1+4
]=[−1,
2
3
]
∴ Co-ordinates of Q=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
−1+5
,
2
4+4
]=[2,4]
∴ Co-ordinates of R=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
5+5
,
2
4−1
]=[5,
2
3
]
∴ Co-ordinates of S=[
2
x
1
+x
2
,
2
y
1
+y
2
]=[
2
5−1
,
2
−1−1
]=[4,−1]
Now, length of PQ=
(−1−2)
2
+(
2
3
−4)
2
=
(−3)
2
+(−
2
5
)
2
=
9+
4
25
=
4
61
Length of QR=
(2−5)
2
+(4−
2
3
)
2
=
(−3)
2
+(
2
5
)
2
=
9+
4
25
=
4
61
Length of RS=
(5−2)
2
+(
2
3
+1)
2
=
(3)
2
+(
2
5
)
2
=
9+
4
25
=
4
61
Length of S=
(2+1)
2
+(−1−
2
3
)
2
=
(3)
2
+(−
2
5
)
2
=
9+
4
25
=
4
61
Length of diagonal PR=
(−1−5)
2
+(
2
3
−
2
3
)
2
=
(6
2
)
=
36
=6
Length of diagonal QS=
(2−2)
2
+(4+1)
2
=
(5)
2
=
25
=5
Hence the all the sides of the quadrilateral PQRS are equal but the diagonals are not equal then PQRS is a rhombus.