Question

If @ and ß are roots of polynomial 4x²+3x+7 then find the value of 1/@ + 1/ß​

Answer 1

Answer:

This is the answer

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Answer 2

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 !