Question

If a and B are the zeroes of the quadratic polynomial f(x)=4x²-5x-1,then find the valueof 1/a + 1/B-aB

Answer 1

Question :-

If a and B are the zeros of the Quadratic polynomial f(x) = 4x² - 5x -1 ,then find the value of -

$\implies \: \sf{ \frac{1}{a} + \frac{1}{B} - aB}$

Answer :-

$\rightarrow \boxed{ \sf{ \frac{1}{a} + \frac{1}{B} - aB} \: = \frac{ - 19}{4} } \:$

Formula used :-

$\: \star \: \: \sf{sum \: of \: zeros = \frac{ - b}{a} } \\ \\ \star \: \: \sf{product \: of \: zeros = \frac{c}{a} }$

Explanation :-

Given equation is

→ 4x² - 5x -1 = 0

if a and B are the zeros of this equation ,

then ,

→ it's sum of zeros = a + B

and

→ product of zeros = aB

according to the given formula

$\sf{sum \: of \: zeros } \downarrow \: \\ \implies \sf{ a + B = \frac{ - coefficient \: of \: {x} }{coefficient \: of \: {x}^{2} } } \\ \\ \implies \sf{a + B = \frac{ - ( - 5)}{4} } \\ \\ \implies \sf{ a + B = \frac{5}{4} } \: \: ....eq.(1) \\ \\ \: \sf{ product \: of \: zeros} \: \downarrow \: \\ \\ \implies \sf{ aB = \frac{constant}{coefficient \: of \: {x}^{2} } } \\ \\ \implies \sf{ aB = \frac{ - 1}{4} } \: ....eq.(2)$

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Now required solution -

$\rightarrow \sf{ \frac{1}{a} + \frac{1}{B} - aB} \\ \\ \rightarrow \: \sf{ \frac{a + B}{aB} - aB} \\ \\ \sf{from \: eq.(1) \: and \: eq.(2)} \\ \\ \rightarrow \: \sf{ \frac{ \frac{5}{4} }{ \frac{ - 1}{4} } - \frac{( - 1)}{4} } \\ \\ \rightarrow \: \sf{ - 5 + \frac{1}{4} } \\ \\ \rightarrow \: \sf{ \frac{ - 20 + 1}{4} } \\ \\ \rightarrow \: \sf{ \frac{ - 19}{4} } \\ \\ hence \\ \\ \rightarrow \boxed{ \sf{ \frac{1}{a} + \frac{1}{B} - aB} \: = \frac{ - 19}{4} }$

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