Question

In a fraction, if the numerator is increased by 2 and the denominator is
decreased by 3, then the fraction becomes 1. Instead, if the numerator is
decreased by 2 and the denominator is increased by 3, then the fraction
becomes3/8

. Find the fraction.

Answer 1

a−b,a,a+b are zeroes of x

a−b,a,a+b are zeroes of x 3

a−b,a,a+b are zeroes of x 3 −3x

a−b,a,a+b are zeroes of x 3 −3x 2

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1⇒b

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1⇒b 2

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1⇒b 2 =2

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1⇒b 2 =2⇒b=±

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1⇒b 2 =2⇒b=± 2

a−b,a,a+b are zeroes of x 3 −3x 2 +x+1⇒a−b+a+a+b= sum of zeroes = 3⇒3a=3⇒a=1(a−b)(a)(a+b)= product of zeroes =−1⇒(1−b)(1+b)=−1⇒1−b 2 =−1⇒b 2 =2⇒b=± 2

Answer 2

Answer:

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