Dear Student, Kx2-x+x+5=0Kx2+1-Kx+5=0x2+1-KKx+5K=0α and β are roots of the equationhence α+β=-baα+β=-1-KK --1and α×β=5K --2Also given thatαβ+βα=45α2+β2α×β=45α2+β2=45×α×βas we know α2+β2=α+β2-2αβhence α+β2-2αβ=45×α×βputting values from equation 1 and 2K-1K2-10K=45×5KK-1K2-10K=4KK-12-10K=4KK2+1-2K-10K-4K=0K2-16K+1=0finding rootsK=16±2522K1=8+63 and K2=8-63Value of K1K2+K2K1=8+63 8-63+8-63 8+63⇒8+63 2+8-63 264-63=64+63+1663+64+63-1663=64+63+64+63=254 Answer is 254 Regards,

Question 7 please

Dear Student,

K (x^{2} - x) + x + 5 = 0 K x^{2} + (1 - K) x + 5 = 0 x^{2} + \frac{1 - K}{K} x + \frac{5}{K} = 0 α a n d β a r e r o o t s o f t h e e q u a t i o n h e n c e α + β = \frac{- b}{a} α + β = \frac{- (1 - K)}{K} - - (1) a n d α \times β = \frac{5}{K} - - (2) A l s o g i v e n t h a t \frac{α}{β} + \frac{β}{α} = \frac{4}{5} \frac{α^{2} + β^{2}}{α \times β} = \frac{4}{5} α^{2} + β^{2} = \frac{4}{5} \times α \times β a s w e k n o w α^{2} + β^{2} = {(α + β)}^{2} - 2 α β h e n c e {(α + β)}^{2} - 2 α β = \frac{4}{5} \times α \times β p u t t i n g v a l u e s f r o m e q u a t i o n 1 a n d 2 {(\frac{K - 1}{K})}^{2} - \frac{10}{K} = \frac{4}{5} \times \frac{5}{K} {(\frac{K - 1}{K})}^{2} - \frac{10}{K} = \frac{4}{K} {(K - 1)}^{2} - 10 K = 4 K K^{2} + 1 - 2 K - 10 K - 4 K = 0 K^{2} - 16 K + 1 = 0 f i n d i n g r o o t s K = \frac{16 \pm \sqrt{252}}{2} K_{1} = 8 + \sqrt{63} a n d K_{2} = 8 - \sqrt{63} V a l u e o f \frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}} = \frac{8 + \sqrt{63}}{8 - \sqrt{63}} + \frac{8 - \sqrt{63}}{8 + \sqrt{63}} \Rightarrow \frac{{(8 + \sqrt{63})}^{2} + {(8 - \sqrt{63})}^{2}}{64 - 63} = 64 + 63 + 16 \sqrt{63} + 64 + 63 - 16 \sqrt{63} = 64 + 63 + 64 + 63 = 254

Answer is 254

Regards,

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