√3x2−7x−30−√2x2−7x−5=x−5
√3x2−7x−30=x−5+√2x2−7x−5
Squaring on both sides,
[√3x2−7x−30]2=[x−5+√2x2−7x−5]2
⇒3x2−7x−30=(x−5)2+2x2−7x−5+2(x−5)√(2x2−7x−5)
⇒3x2−7x−30=x2−10x+25+2x2−7x−5+2(x−5)√(2x2−7x−5)
⇒3x2−7x−30=3x2−17x+20+2(x−5)√(2x2−7x−5)
⇒−7x+17x−30−20=2(x−5)√(2x2−7x−5)
⇒10x−50=2(x−5)√(2x2−7x−5)
⇒10(x−5)=2(x−5)√(2x2−7x−5)
⇒5=√(2x2−7x−5)
Squaring on both sides
⇒25=2x2−7x−5
⇒2x2−7x−5−25=0
⇒2x2−7x−30=0
⇒(2x+5)(x−6)=0 on factorising
∴x=−52,6
For x=−52 the √3x2−7x−30−√2x2−7x−5=x−5 equation becomes
√3(−52)2−7×−52−30−√2(−52)2−7×−52−5=−52−5
=√754+352−30−√2×254+352−5
=√145−1204−√402
=√254−√20=52−4√5≠ RHS
∵x−5=−52−5=−152
Hence x=−52 is not a solution
For x=6 the √3x2−7x−30−√2x2−7x−5=x−5 equation becomes
√3(6)2−7×6−30−√2(6)2−7×6−5=6−5
=√108−72−√72−47=1
=√36−√25=6−5=1
Thus, x=6 is the only solution for the above equation