If a+b are zero of polynomial p(x)= x²-5x+k such that a-b = 1 find the value of k
Answers
Answer:
k = 6
Note:
★ The possible values of the variable for which the polynomial becomes zero are called its zeros .
★ A quadratic polynomial can have atmost two zeros .
★ If a and b are the zeros of the quadratic polynomial Ax² + Bx + C , then ;
• Sum of zeros , (a + b) = -B/A
• Product of zeros , (ab) = C/A
Solution:
Here,
The given quadratic polynomial is ;
p(x) = x² - 5x + k .
Comparing with the general form of a quadratic polynomial Ax² + Bx + C ,
We have ;
A = 1
B = -5
C = k
Also,
It is given that , a and b are the zeros of the given quadratic polynomial p(x) .
Thus,
=> Sum of zeros = -B/A
=> a + b = -(-5)/1
=> a + b = 5 ---------(1)
Also,
It is given that ;
a - b = 1 -----------(2)
Now,
Adding eq-(1) and (2) , we get ;
=> a + b + a - b = 5 + 1
=> 2a = 6
=> a = 6/2
=> a = 3
Now,
Putting a = 3 in eq-(1) , we get ;
=> a + b = 5
=> 3 + b = 5
=> b = 5 - 3
=> b = 2
Now,
=> Product of zeros = C/A
=> a × b = k/1
=> 3 × 2 = k
=> k = 6
Hence , k = 6
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Another method :
Here,
The given quadratic polynomial is ;
p(x) = x² - 5x + k .
Comparing with the general form of a quadratic polynomial Ax² + Bx + C ,
We have ;
A = 1
B = -5
C = k
Also,
It is given that , a and b are the zeros of the given quadratic polynomial p(x) .
Thus,
=> Sum of zeros = -B/A
=> a + b = -(-5)/1
=> a + b = 5 ---------(1)
Also,
=> Product of zeros = C/A
=> ab = k/1
=> ab = k ------------(2)
Also,
It is given that ;
a - b = 1 -----------(3)
Also,
We know that ;
=> (a + b)² = (a - b)² + 4ab
{ using eq-(1) , (2) and (3) }
=> 5² = 1² + 4k
=> 25 = 1 + 4k
=> 25 - 1 = 4k
=> 24 = 4k
=> k = 24/4
=> k = 6