If P = a 2 − b 2 + 2ab, Q = a 2 + 4b 2 − 6ab, R = b 2 + b, S = a 2 − 4ab and T = −2a 2 + b 2 – ab + a. Find P + Q + R + S –t
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Answer:
Given
P=a
2
−b
2
+2ab
Q=a
2
+4b
2
−6ab
R=b
2
+b
S=a
2
−4ab and
T=−2a
2
+b
2
−ab+a
∴P+Q+R+S−T
=(a
2
−b
2
+2ab)+(a
2
+4b
2
−6ab)+(b
2
+b)+(a
2
−4ab)−(−2a
2
+b
2
−ab+a)
=a
2
−b
2
+2ab+a
2
+4b
2
−6ab+b
2
+b+a
2
−4ab+2a
2
−b
2
+ab−a
Rearranging and collecting like terms,
=a
2
+a
2
+a
2
+2a
2
−b
2
+4b
2
+b
2
−b
2
+2ab−6ab−4ab+ab+b−a
=(1+1+1+2)a
2
+(−1+4+1−1)b
2
+(2−6−4+1)ab+b−a
=5a
2
+3b
2
−7ab+b−a
Hence, P+Q+R+S−T=5a
2
+3b
2
−7ab+b−a
Step-by-step explanation:
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