Explanation
Given quadratic equation is √7y2−6y−13√7=0
Comparing it with the standard form of equation ax2+bx+c we get a=7,b=6,c=13√7 Therefore, x=−b±√b2−4ac2a=−(−6)±√(−6)2−4×√7×(−13√7)2√7=6±√36+3642√7=6±√4002√7=6±202√7=6−202√7=−142√7=−7√7=−7×√7√7×√7=−√7=6+202√7=262√7=13√7
The quadratic equation is 2x2−2√2x+1=0.Comparing it with standard form of quadratic equation ax2+bx+c=0,we get a=2,b=−2√2,c=1 Therefore the roots of the quadratic equation are,x=−b±√b2−4ac2a =−(−2√2)±√(−2√2)2−4×2×12×2 =(2√2)±√8−84 =2√2±04 =2√24 =√22 =1√2
Given, xx−1+x−1x=4⇒x2+(x−1)2x(x−1)=4⇒x2+x2−2x+1=4x2−4x⇒2x2−4x2−2x+4x+1=0⇒−2x2+2x+1=0⇒2x2−2x−1=0Comparing 2x2−2x−1=0 with ax2+bx+c=0, we get a=2,b=−2.c=−1Now the roots are = −b±√b2−4ac2a= −(−2)±√(−2)2−4×2×(−1)2×1= 2±√4+82= 2±√122= 2±2√32= 1±√3
Given, x+1x=3⇒x+1x=3⇒x2−3x+1=0Comparing x2−3x+1=0 with ax2+bx+c=0, we get a=1,b=3.c=14Now the roots are = −b±√b2−4ac2a= −(−3)±√(3)2−4×1×12×1= 3±√9−42= 3±√52= 3+√52 or 3−√52
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