y-10y^0.5 + 23 = 0
using answer b (x^2 -10x + 23) = x=root2 + 5 / x=-root2 +5
write the solution in form p+q root r where p, q, r are integers
write the solution in form p+q root r where p, q, r are integers
Original post by cheery-condition
using answer b (x^2 -10x + 23) = x=root2 + 5 / x=-root2 +5
write the solution in form p+q root r where p, q, r are integers
write the solution in form p+q root r where p, q, r are integers
What have you tried?
Original post by cheery-condition
using answer b (x^2 -10x + 23) = x=root2 + 5 / x=-root2 +5
write the solution in form p+q root r where p, q, r are integers
write the solution in form p+q root r where p, q, r are integers
Recognise the fact that it’s a quadratic. But instead of having x, it’s root y. because y = (root y)^2 right. and y^0.5 = root y. so you can say x=sqrt y, then solve the quadratic for x, then since your answers are for root y, you square your answers to get y.
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