Question

plz answer its urgent question 5 and 6

Answer 1

here is your answer mate ❤️
Hope it helps you

Answer 2

Answer of 5th question is Option (a) which is equal to 2. Answer of 6th question is option (c) which is plus minus 2.

1: The 5th question surely has a printing mistake. If there wasn't a mistake, the first part would have been a/c instead of a^2/ac. My answer matches with one option and I think that must be the correct thing to prove.
Let
$\alpha \: and \: \beta$
be the roots of quadratic equation
$a {x}^{2} + bx + c = 0$
Sum of roots of quadratic equation =

Sum of Squares of their reciprocals.

$\alpha + \beta = \frac{1}{ { \alpha }^{2} } + \frac{1}{ \beta ^{2} } \\ \alpha + \beta = \frac{ { \alpha }^{2} + { \beta }^{2} }{ { \alpha }^{2} { \beta }^{2} } \\ \alpha + \beta = \frac{( \alpha + \beta ) ^{2} - 2 \alpha \beta }{( \alpha \beta )^{2} }$
Sum of roots = -b/a
Product of roots = c/a

$\alpha + \beta = \frac{( \alpha + \beta ) ^{2} - 2 \alpha \beta }{( \alpha \beta )^{2} } \\ \frac{ - b}{a} = \frac{ ( - \frac{b}{a}) ^{2} - 2( \frac{c}{a} ) }{ ({ \frac{c}{a} )}^{2} } \\ \frac{ - b}{a} = \frac{ \frac{ {b}^{2} }{ {a}^{2} } - \frac{2c}{a} }{ \frac{ {c}^{2} }{ {a}^{2} } } \\ \frac{ - b}{a} = \frac{ {b}^{2} - 2ac }{ {a}^{2} } \times \frac{ {a}^{2} }{ {c}^{2} } \\ \frac{ - b}{a} = \frac{ {b} ^{2} - 2ac }{ {c}^{2} } \\ 0 = \frac{ {b^{2} - 2ac}}{ {c}^{2} } + \frac{b}{a} \\ 0 = \frac{ {b}^{2} }{ {c}^{2} } - \frac{2a}{c} + \frac{b}{a} \\ 0 = \frac{a {b}^{2} - 2 {a}^{2} {c} + b {c}^{2} }{ac ^{2} } \\ 0 = a {b}^{2} - 2 {a}^{2} c + b {c}^{2} \\ 0 = \frac{a {b}^{2} }{a ^{2}c } - \frac{2a ^{2}c }{a ^{2}c } + \frac{b {c}^{2} }{a ^{2} c} \\ 2= \frac{ {b}^{2} }{ac} + \frac{bc}{ {a}^{2} } \:$

2:

${x}^{2} - (p + 4)x + 2p + 5 = 0$
has equal roots.

Therefore,

$(p + 4)^{2} - 4(2p + 5) = 0 \\ {p}^{2} + 8p + 16 - 8p - 20 = 0 \\ {p}^{2} - 4 = 0 \\ p = \pm 2$