Question

If x and y are the zeroes of the polynomial t2 – t + 4 find the value of (1/x) +(1/y) –xy.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{(\frac{1}{x})+(\frac{1}{y})-xy=\frac{-15}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green {\underline \bold{given: }} \\ \tt{: \implies {t}^{2} - t + 4 = 0} \\ \\ \tt{: \implies x \: and \: y \: are \: zeroes \: of \: this \: quadratic} \\ \\ \red {\underline\bold{to \: find: }} \\ \tt{: \implies \frac{1}{x} + \frac{1}{y} - xy = ?}$

• According to given question :

$\tt{:\implies {t}^{2} - t + 4 = 0}\\ \\ \bold{solving \: this \: quadratic \: by \:qudratic \: formula} \\ \tt{: \implies t = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2} } \\ \\ \tt{: \implies t = \frac{ - ( - 1) \pm \sqrt{ {( - 1) }^{2} - 4 \times 1 \times 4} }{2 \times 1} } \\ \\ \tt{: \implies t = \frac{1 \pm \sqrt{1 - 16} }{2}} \\ \\ \tt{: \implies t = \frac{1 \pm \sqrt{ - 15} }{2}} \\ \\ \tt{: \implies t = \frac{1 \pm15 \iota}{2} } \\ \\ \green{ \tt{\therefore x = \frac{1 + 15 \iota}{2}}} \\ \\ \green{ \tt{\therefore y = \frac{1 - 15 \iota}{2}}} \\ \\ \bold{for \: finding \: value} \\ \tt{: \implies \frac{1}{x} + \frac{1}{y} - xy } \\ \\ \text{putting \: given \: values}\\ \tt{: \implies \frac{1}{ \frac{1 + 15 \iota}{2} } + \frac{1}{ \frac{1 - 15 \iota}{2} } - \frac{1 + 15 \iota}{2} \times \frac{1 - 15 \iota}{2} } \\ \\ \tt{: \implies \frac{2}{1 + 15 \iota} + \frac{2}{1 - 15 \iota} - \frac{ {1}^{2} - {(15 \iota)}^{2} }{4} } \\ \\ \tt{: \implies \frac{2(1 - 15 \iota) + 2(1 + 15 \iota}{(1 + 15 \iota)(1 - 15 \iota)} - \frac{1 +15 }{4} } \\ \\ \tt{: \implies \frac{2 - 30 \iota + 2 + 30 \iota}{1 + 15} - 4} \\ \\ \tt{: \implies \frac{2 + 2}{16} - 4} \\ \\ \tt{: \implies \frac{1}{4} - 4} \\ \\ \green{\tt{: \implies \frac{ - 15}{4} }}$

Answer 2

Answer

-15/4

$\rule{200}2$

Explanation

Given, x and y are the zeroes of polynomial t² - t + 4.

To find, the value of (1/x) + (1/y) - xy

Comparing the given quadratic equation with standard equation ax² + bx + c, we get

a = 1

b = -1

c = 4

We know that, for a quadratic equation, sum of zeroes = -b/a

⇒ x + y = - (-1/1)

⇒ x + y = 1.........(1)

And, product of zeroes = c/a

⇒ xy = 4/1

⇒ xy = 4.................(2)

Now,

(1/x) + (1/y) - xy

Taking LCM,

⇒ (x + y)/xy - xy

From (1) and (2), the expression becomes

⇒ 1/4 - 4

⇒ 1/4 - 16/4

⇒ -15/4