the product of 2 consecutive multiple of 7 is 768.write statement in algebra
Answers
Let two consecutive numbers be xx and x+7x+7
According to the question,
\Rightarrow⇒ x^2+(x+7)^2=637x
2
+(x+7)
2
=637
\Rightarrow⇒ x^2+x^2+14x+49=637x
2
+x
2
+14x+49=637
\Rightarrow⇒ 2x^2+14x+49-637=02x
2
+14x+49−637=0
\Rightarrow⇒ 2x^2+14x-588=02x
2
+14x−588=0
\Rightarrow⇒ 2(x^2+7x-294)=02(x
2
+7x−294)=0
\Rightarrow⇒ x^2+7x-294=0x
2
+7x−294=0
\Rightarrow⇒ x^2+21x-14x-294=0x
2
+21x−14x−294=0
\Rightarrow⇒ x(x+21)-14(x+21)=0x(x+21)−14(x+21)=0
\Rightarrow⇒ (x+21)(x-14)=0(x+21)(x−14)=0
\Rightarrow⇒ x+21=0x+21=0 and x-14=0x−14=0
\Rightarrow⇒ x=-21x=−21 and x=14x=14
\Rightarrow⇒ Here, one number is 1414
\Rightarrow⇒ Other, number =x+7=14+7=21=x+7=14+7=21
\Rightarrow⇒ The product of two multiple 14\times 21=29414×21=294
hope it helps
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Answer:
not knowing sorry....