Question

please answer my questions tennetiraj sir please​

Answer 1

Step-by-step explanation:

let the zeros be x1 and x2

x1+x2=a

x1-x2=1

x1x2=b

(x1+x2)^2=(x1-x2)^2+4x1x2

a^2=1+4b

option A

Answer 2

Answer:

Option (A)

Step-by-step explanation:

Given :-

The zeroes of x^2-ax+b are differ by unity

To find :-

Which one of the following is true ?

Solution :-

Given Polynomial = x^2-ax+b

On Comparing this with the standard quadratic Polynomial ax^2+bx+c

We have, a = 1 , b= -a ,c = b

Let the zeroes are A and B

Given that

They differ by unity

A- B = 1

=> A = B+1 ----------(1)

We know that

Sum of the zeroes = -b/a

=> A+ B = -(-a)/1

=> B+1+B = a

=> 2B+1 = a

=> 2B = a-1

=> B = (a-1)/2---------(2)

Product of the zeroes = c/a

=> AB = b/1

=> AB = b

=> (B+1)(B) = b

=> [{(a-1)/2}+1](a-1)/2 = b

=> [(a-1+2)/2][(a-1)/2] = b

=> [(a+1)/2][(a-1)/2] = b

=> (a+1)(a-1)/4 = b

=> (a^2-1^2)/4 = b

Since (a+b)(a-b)=a^2-b^2

=> (a^2-1)/4 = b

=> a^2-1 = 4×b

=> a^2 -1 = 4b

=> a^2 = 4b+1

=> a^2 = 1+4b

Answer:-

If the zeroes are differ by unity of x^2-ax+b then a^2=1+4b is true.

Used formulae:-

• The standard quadratic Polynomial ax^2+bx+c

• Sum of the zeroes = -b/a

• Product of the zeroes = c/a

• (a+b)(a-b)=a^2-b^2