Under rootx+5 +under root x+21= under root 6x+40
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Answer:
How will you solve the equation √(x+5) + √ (x+21) = √ (6x+40)?
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x+5−−−−−√+x+21−−−−−√=6x+40−−−−−−√
Square both sides:
(x+5)+(x+21)+2(x+5)(x+21)−−−−−−−−−−−−√=6x+40
2x+26+2(x+5)(x+21)−−−−−−−−−−−−√=6x+40
2(x+5)(x+21)−−−−−−−−−−−−√=4x+14
(x+5)(x+21)−−−−−−−−−−−−√=2x+7
Square both sides again:
(x+5)(x+21)=(2x+7)2
x2+26x+105=4x2+28x+49
3x2+2x−56
3x2−12x+14x−56
3x(x−4)+14(x−4)
(x−4)(3x+14)
x=4,x=−143
Check for extraneous solutions that might have been introduced by squaring:
x=4
4+5−−−−√+4+21−−−−−√=9–√+25−−√=3+5=8
6(4)+40−−−−−−−−√=24+40−−−−−−√=64−−√=8
x=−143
−143+5−−−−−−−√+−143+21−−−−−−−−√=13−−√+493−−−√=83–√
6(−143)+40−−−−−−−−−−−√=−28+40−−−−−−−−√=12−−√=23–√
So the only valid solution is:
x=4
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Let us attempt analytical solution.
Squaring both the sides, we get
x+5 +sqrt((x+5)(x+21))+x+21=6x+40
=> 2x +26+2sqrt(x^2+26x+105)=6x + 40
=> 2sqrt(x^2+26x+105)=4x + 14
=> sqrt(x^2+26x+105)=2x + 7
Again squaring both the sides, we get
x^2+26x+105=(2x + 7)^2
=> x^2+26x+105=4x^2+14x+49
=> 0=3(x^2)+2x+56
Multiplying both sides by 3, we get
0=9(x^2)+6x+168
=>(3x)^2+2(3x)(1)+1–169=0
=> (3x+1)^2-(13^2)=0
=>(3x+1-13)(3x+1+13)=0
=>(3x-12)(3x+14)=0
=>3x=12 or 3x=-14
=>x=4 or x=-14/3
For x=4, Equation is trivially satisfied.
For 14/3, first term becomes sqrt(0.3333); second term becomes sqrt(16.3333) and term on right sid
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How will you solve the equation √ (3x +1) - √(x-1) = 2?
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