Question : If x=1/(7-4√3), show that x²-14x+1=0 and hence evaluate x³-9x²-69x+7.
Doubt by Purvi
Solution :
x=1/(7-4√3)
Rationising the denominator
x=1/(7-4√3)×(7+4√3)/(7+4√3)
x=(7+4√3)/[(7)²-(4√3)²]
x=(7+4√3)/(49-48)
x=(7+4√3)/1
x=7+4√3
x²-14x+1=0
LHS
x²-14x+1
=[7+4√3]²-14[7+4√3]+1
=(7)²+(4√3)²+2(7)(4√3)-14(7)-14(4)√3+1
=49+48+56√3-98-56√3+1
=50+48-98
=98-98
=0
= RHS
LHS = RHS
Hence Proved.
Now
x²-14x+1=0
x²=14x-1— (1)
Also
x³=x².x
x³=(14x-1)x [∵ Using Equation (1)]
x³=14x²-x — (2)
Now
x³-9x²-69x+7
=14x²-x-9(14x-1)-69x+7
[Using Equation (1) and (2)]
=14x²-x-126x+9-69x+7
=14x²-196x+16
=14x²-196x+14+2
=[14x²-196x+14]+2
=14[x²-14x+1]+2
=14[0]+2 [∵x²-14x+1=0]
=0+2
=2
Hence, x³-9x²-69x+7=2