Question
Which of the following is a zero of (49x2−1)+(1+7x)2 ?
Answer: Option A
:
A
Given,(49x2−1)+(1+7x)2=((7x)2−12)+(1+7x)2=(7x−1)(7x+1)+(1+7x)(1+7x)=(1+7x)(7x−1+1+7x)[Taking(1+7x)common]=(1+7x)(14x)
To find the zeroes of a polynomialexpression,we must equate it to zero.(1+7x)(14x)=0⇒(1+7x)=0;(14x)=0∴x=−17,0Zeroes are−17&0.
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:
A
Given,(49x2−1)+(1+7x)2=((7x)2−12)+(1+7x)2=(7x−1)(7x+1)+(1+7x)(1+7x)=(1+7x)(7x−1+1+7x)[Taking(1+7x)common]=(1+7x)(14x)
To find the zeroes of a polynomialexpression,we must equate it to zero.(1+7x)(14x)=0⇒(1+7x)=0;(14x)=0∴x=−17,0Zeroes are−17&0.
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