Question

5. If a and B are the zeroes of the quadratic polynomial f(x) = x - x - 4, then the value of 1/a + 1/b - ab, is
(a) 15/4
(b) -15/4
(c) 4
(d) 15​

Answer 1

Answer:

${\bold{\therefore \frac{1}{a}+\frac{1}{b}-ab=\frac{15}{4}}}$

Step-by-step explanation:

$\huge{\bold{\underline{\underline{SOLUTION-}}}}$

• In the given question information given about a quadratic whose zeroes are a and b.

• We have to find the given value.

$\underline \bold{Given : } \\ \implies a \: and \: b \: \in ({x}^{2} - x -4 ) \\ \\ \underline \bold{To \: Find : } \\ \implies \frac{1}{a} + \frac{1}{b} - ab =?$

• According to given question :

$\bold{Quadratic \: formula : } \\ \implies {x}^{2} - x - 4 = 0 \\ \\ \implies x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \implies x = \frac{ - ( - 1) + \sqrt{ (- 1)^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} \\ \\ \bold{\implies x = \frac{1 + \sqrt{17} }{2} - - (first \: zeroes)} \\ \\ \bold{\implies x = \frac{1 - \sqrt{17} }{2} - - (second \: zeroes)} \\ \\ \bold{sum \: of \: zeroes} \\ \implies a + b = \frac{1 + \sqrt{17} }{2} + \frac{1 - \sqrt{17} }{2} \\ \\ \implies a + b = \frac{1 + \cancel{\sqrt{17}} + 1 - \cancel{\sqrt{17}} }{2} \\ \\ \implies a + b = \frac{ \cancel2}{ \cancel2} \\ \\ \bold{\implies a + b = 1} \\ \\ \bold{product \: of \: zeroes} \\ \implies a \times b = \frac{1 + \sqrt{17} }{2} \times \frac{1 - \sqrt{17} }{2} \\ \\ \implies ab = \frac{ {1}^{2} - (\sqrt{17})^{2} }{4} \\ \\ \implies ab = \frac{1 - 17}{4} \\ \\ \implies ab = \frac{ \cancel{- 16}}{ \cancel4} \\ \\ \bold{ \implies ab = - 4 }\\ \\ \bold{finding \: given \: value : } \\ \implies \frac{1}{a} + \frac{1}{b} - ab \\ \\ \implies \frac{a + b}{ab} - ab \\ \\ \implies \frac{1}{ - 4} - ( - 4) \\ \\ \implies \frac{ - 1}{4} + 4 \\ \\ \implies \frac{ - 1 + 16}{4} \\ \\ \implies \bold{\frac{15}{4}}$