The number of real solutions of the equation (15+4√14)t+(15−4√14)t=30, where t=x2−2∣x∣ are
For the equation, (15−4√14)t+(15+4√14)t=30 to be true,
t=±1
Now, t=x2−2|x|
x2−2|x|=±1
x2−2|x|+1=0
(|x|−1)2=0
x=±1 and x2−2|x|−1=0
|x|=2±√4+42
|x|=1±√2
Hence, 4 values.