√x2+3x+9 + √x2+3x+4 =5
Let y=x2+3x
√y+9+√y+4=5
√y+9=5−√y+4
Taking 'square' both sides
(√y+9)2=(5−√y+4)2
y+9=(5)2−2(5)(√y+4)+(√y+4)2
y+9=25−10√y+4+y+4
10√y+4=25+y+4−y−9
10√y+4=20√y+4=2Taking square both sides(√y+4)2=(2)2y+4=4y=4−4y=0Restoring the value of yx2+3x=0x(x+3)=0x=0 x+3=0x=−3CheckFor x=0√(0)2+3(0)+9+√(0)2+3(0)+4=5√0+0+9+√0+0+4=5√9+√4=53+2=55=5For x=−3√(−3)2+3(−3)+9+√(−3)2+3(−3)+4=5√0+0+9+√0+0+4=5√9+√4=53+2=55=5