Average and standard deviation equation system 
Every exam has 100 points. Student's average of 3 exams is 88 points and standard deviation is 3. How many points should he get on fourth and fifth exam so that average is 90 and standard deviation stays same (=3).

I tried solving equations $(x_1+x_2+x_3)/3=88$ and $(x_1+x_2+x_3+x_4+x_5)/5=90$. Since $x_1+x_2+x_3=264$ it means that $(264+x_4+x_5)/5=90$ so $x_4+x_5=186$. Then I tried to use equations for standard deviation but I haven't managed to get anything useful.
 A: We have
$$\sqrt{\frac{(x_1-88)^2+(x_2-88)^2+(x_3-88)^2}3}=3$$
$$(x_1-88)^2+(x_2-88)^2+(x_3-88)^2=27$$
How do we get from $(x-88)^2=x^2-176x+88^2$ to $(x-90)^2=x^2-180x+90^2$? We add $-4x+356$.
$$-4x_1+356-4x_2+356-4x_3+356=-4(x_1+x_2+x_3)+1068=-4(3\cdot88)+1068=12$$
$$(x_1-90)^2+(x_2-90)^2+(x_3-90)^2=27+12=39$$
Now
$$\sqrt{\frac{39+(x_4-90)^2+(x_5-90)^2}5}=3$$
$$(x_4-90)^2+(x_5-90)^2=6$$
Let $z_4=x_4-90$ and $z_5=x_5-90$. Then $z_4+z_5=6$ while $z_4^2+z_5^2=6$. Squaring the first equation and subtracting the second from it and halving gives $z_4z_5=15$.
Thus we may form the polynomial $x^2-6x+15$, whose roots should be $z_4$ and $z_5$. However, we find that the polynomial has no real roots. So there is no solution to the original problem – the average and standard deviation cannot be changed by the further two tests to the given values.
A: From the averages,
$$x_4+x_5=5\cdot 90-3\cdot 88=186$$
and from the variances,
$$x_4^2+x_5^2=5(3^2+90^2)-3(3^2+88^2)=17286.$$
Then
$$(x_4-x_5)^2=2(x_4^2+x_5^2)-(x_4+x_5)^2=-24.$$ (!)
