If α and β are zeros of the polynomial P(x) = 4 x²+ 3 x + 7, then ( 1/α + 1/β) is equal to
Answers
Answer :
1/α + 1/ß = -3/7
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 .
★ The general form of a quadratic polynomial is given as ; ax² + bx + c .
★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;
• Sum of zeros , (α + ß) = -b/a
• Product of zeros , (αß) = c/a
Solution :
Here ,
The given quadratic polynomial is ;
p(x) = 4x² + 3x + 7 .
Comparing the given quadratic polynomial with the general quadratic polynomial ax² + bx + c , we have ;
a = 4
b = 3
c = 7
Also ,
It is given that , α and ß are the zeros of the given quadratic polynomial p(x) .
Thus ,
=> Sum of zeros = -b/a
=> α + ß = -3/4
Also ,
=> Product of zeros = c/a
=> αß = 7/4
Now ,
=> 1/α + 1/ß = (ß + α) / αß
=> 1/α + 1/ß = (α + ß) / αß
=> 1/α + 1/ß = ( -3/4 ) / (:7/4 )
=> 1/α + 1/ß = ( - 3/4 ) × ( 4/7 )
=> 1/α + 1/ß = -3/7
Hence ,
1/α + 1/ß = -3/7
Añswèr :
- -3/7
Note :
- If ax²+bx+c is a polynomial and @ , ß be the zeroes of the polynomial , then ,
- @ + ß = -b/a
- @ß = c/a
- 1/@ + 1/ß = -b/c
Formulà used :
- 1/@ + 1/ß = -b/c
Solution :
- Comparing the given polynomial with standard form ,
- a = 4 , b = 3 , c = 7
- 1/@ + 1/ß = -3/7
Proof :
- Let ax²+bx+c be a quadratic polynomial , equating the polynomial to zero , we get zeroes
- Let ax²+bx+c = 0 be a quadratic equation, and @ , ß be the roots of the equation ( i.e zeroes of the polynomial)
- Dividing by a on both sides ,
- x² + (b/a)x + c/a = 0
- Transposing constant to RHS
- x² + (b/a)x = - c/a
- multiplying and dividing by 2 for the term "(b/a)x"
- x² + 2(b/2a)x = -c/a
- adding b²/4a² on both sides
- x² + 2(b/2a)x + b²/4a² = b²/4a² - c/a
- LHS is in the form (A+B)² = A²+B²+2AB , where , A = x and B = b/2a
- (x+b/2a)² = (b²-4ac)/4a²
- Applying square root on both sides ,
- x + b/2a = ±√(b²-4ac)/√(2a)²
- x + b/2a = ± √(b²-4ac)/2a
- x = {-b/2a} ± {√(b²-4ac)/2a}
- x = -b±√b²-4ac/2a
- x = -b±√∆/2a , where ∆ = b²-4ac
- x = (-b+√∆) / 2a and x = (-b-✓∆) / 2a are zeroes of the polynomial
- Therefore , @ = (-b+✓∆) / 2a and ß=(-b-✓∆) / 2a
Consider @ + ß :
- @ + ß = (-b+✓∆) / 2a + (-b-✓∆) / 2a
- @ + ß = (-b + √∆ -b -√∆ ) / 2a
- @ + ß = -2b / 2a
- @ + ß = - b/a ----(i)
Consider @ß :
- @ß = (-b+✓∆) / 2a * (-b-✓∆) / 2a
- @ß = (-b+✓∆)(-b-✓∆)/4a²
- Numerator is in form (A+B)(A-B) = A²-B² , where A = -b and B = √∆
- @ß = (-b)²-(√∆)² /4a²
- @ß = b² - ∆ / 4a²
- We , know , ∆ = b²-4ac , substituting it ,
- @ß = b²-(b²-4ac)/4a²
- @ß = b² - b² + 4ac / 4a²
- @ß = 4ac/4a²
- @ß = c/a ----(ii)
Consider 1/@ + 1/ß :
- 1/@ + 1/ß
- By , fraction addition ,
- 1/@ + 1/ß = ß + @ / @ß
- 1/@ + 1/ß = @+ß/@ß
- 1/@ + 1/ß = (-b/a) / (c/a) [since , equations (i) and (ii)
- 1/@ + 1/ß = -b/c
Hope it helps. !
- Sorry for the inconvenience , I've used "@" in place of "alpha" and "ß" in place of "beta"