Question

k+5 P k+1=11(k-1)÷2×k+3 P k find k

Answer 1

Given equatin is

$k+5Pk+1=\frac{11(k-1)}{2}*k+3Pk$

multiply by 2 to both sides

2(k+5Pk+1)=2(\frac{11(k-1)}{2}*k)+2(3Pk)

2k+10Pk+2=11(k-1)k+6Pk

2k+10Pk-6Pk+2=11(k-1)k

$2k+4Pk+2=11(k^2-k)$

$2k+4Pk+2=11k^2-11k$

$2k+4Pk+2-11k^2+11k=0$

$4Pk+2-11k^2+13k=0$

$-4Pk-2+11k^2-13k=0$

$11k^2-4Pk-2-13k=0$

$11k^2+(-4P-13)k-2=0$

Now it looks like a quadratic equation $ax^2+bx+c=0$

where a=11, b=(-4P-13), c=-2

so we can plug values of a,b,c into quadratic formula to find value of k

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

plug the values of a, b, c

$k=\frac{-(-4P-13) \pm \sqrt{(-4P-13)^2-4(11)(-2)}}{2(11)}$

$k=\frac{4P+13 \pm \sqrt{(-4P-13)^2+88}}{22}$

$k=\frac{4P+13 \pm \sqrt{(4P+13)^2+88}}{22}$

$k=\frac{4P+13 \pm \sqrt{16P^2+104P+169+88}}{22}$

$k=\frac{4P+13 \pm \sqrt{16P^2+104P+257}}{22}$

It can't be simplified any more because we don't know value of P

Hence final answer is $k=\frac{4P+13 \pm \sqrt{16P^2+104P+257}}{22}$.