If one zero of quadric polynomial x square +(p+1)x-6 is 2,then what is the value of P?
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Answer:
If one zero of quadric polynomial x square +(p+1)x-6 is 2,then what is the value of P?
Let the given polynomial be
One of the zeros of the polynomial is 2
NoTE
Sum of Zeros : - x coefficient/x² coefficient
Product of Zeros : constant term /x² coefficient
Let the other zero be a
Here,
The zeros of the polynomial are 2 and - 3
Also,
Answered by
11
Solution
Given :-
- Polynomial equation, x² + (p+1)x - 6 = 0
- First zero be 2
Find :-
- Value of p
Explanation
We Know,
If 2 is zeros of this polynomial,
Its means that 2 is satisfied this polynomial.
So, Keep x = 2 in this polynomial,
➡ (2)² + (p+1) × 2 - 6 = 0
➡4 + 2(p+1) - 6 = 0
➡2(p + 1) = 6 - 4
➡2(p+1) = 2
➡ (p + 1) = 2/2
➡(p+1) = 1
➡p = 1 - 1
➡p = 0
Hence
- Value of p will be 0
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Answer Verification
keep value of p & x = 2 ,in given polynomial,
➡ (2)² + (0+1)×2 - 6 = 0
➡ 4 + 1 × 2 - 6 = 0
➡ 4 + 2 - 6 = 0
➡ 6 - 6 = 0
➡0 = 0
L.H.S. = R.H.S.
That's Proved.
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