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Question Number 22313 by math solver last updated on 15/Oct/17

Commented by Joel577 last updated on 16/Oct/17

(√(5x + 7)) − (√(3x + 1)) = (√(x + 3)) x ≥ −(7/5) and x ≥ −(1/3) and x ≥ −3 ⇒ x ≥ −(1/3) ((√(5x + 7)) − (√(3x + 1)))^2 = x + 3 5x + 7 − 2(√((5x + 7)(3x +1))) + 3x +1 = x + 3 7x + 5 − 2(√((5x + 7)(3x +1))) = 0 7x + 5 = 2(√((5x + 7)(3x + 1))) 49x^2 + 70x + 25 = 60x^2 + 104x + 28 11x^2 + 34x + 3 = 0 x_1 = −3 x_2 = −(1/(11)) Hence, the solution that satisfy the equation above is {−(1/(11))}

$$\sqrt{\mathrm{5}{x}\:+\:\mathrm{7}}\:−\:\sqrt{\mathrm{3}{x}\:+\:\mathrm{1}}\:=\:\sqrt{{x}\:+\:\mathrm{3}} \\ $$$${x}\:\geqslant\:−\frac{\mathrm{7}}{\mathrm{5}}\:\:\mathrm{and}\:\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{and}\:\:{x}\:\geqslant\:−\mathrm{3} \\ $$$$\Rightarrow\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$\left(\sqrt{\mathrm{5}{x}\:+\:\mathrm{7}}\:−\:\sqrt{\mathrm{3}{x}\:+\:\mathrm{1}}\right)^{\mathrm{2}} \:=\:{x}\:+\:\mathrm{3} \\ $$$$\mathrm{5}{x}\:+\:\mathrm{7}\:−\:\mathrm{2}\sqrt{\left(\mathrm{5}{x}\:+\:\mathrm{7}\right)\left(\mathrm{3}{x}\:+\mathrm{1}\right)}\:+\:\mathrm{3}{x}\:+\mathrm{1}\:=\:{x}\:+\:\mathrm{3} \\ $$$$\mathrm{7}{x}\:+\:\mathrm{5}\:−\:\mathrm{2}\sqrt{\left(\mathrm{5}{x}\:+\:\mathrm{7}\right)\left(\mathrm{3}{x}\:+\mathrm{1}\right)}\:=\:\mathrm{0} \\ $$$$\mathrm{7}{x}\:+\:\mathrm{5}\:=\:\mathrm{2}\sqrt{\left(\mathrm{5}{x}\:+\:\mathrm{7}\right)\left(\mathrm{3}{x}\:+\:\mathrm{1}\right)} \\ $$$$\mathrm{49}{x}^{\mathrm{2}} \:+\:\mathrm{70}{x}\:+\:\mathrm{25}\:=\:\mathrm{60}{x}^{\mathrm{2}} \:+\:\mathrm{104}{x}\:+\:\mathrm{28} \\ $$$$\:\:\mathrm{11}{x}^{\mathrm{2}} \:+\:\mathrm{34}{x}\:\:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$$${x}_{\mathrm{1}} \:=\:−\mathrm{3} \\ $$$${x}_{\mathrm{2}} \:=\:−\frac{\mathrm{1}}{\mathrm{11}} \\ $$$$ \\ $$$$\mathrm{Hence},\:\mathrm{the}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{above}\:\mathrm{is}\:\:\left\{−\frac{\mathrm{1}}{\mathrm{11}}\right\} \\ $$

Commented by math solver last updated on 16/Oct/17

thanks:)!

$$\left.{thanks}:\right)! \\ $$

Answered by squidward last updated on 16/Oct/17

it is a quadratic equation, hence there are 2 solutions

$${it}\:{is}\:{a}\:{quadratic}\:{equation},\:{hence} \\ $$$${there}\:{are}\:\mathrm{2}\:{solutions} \\ $$

Commented by squidward last updated on 16/Oct/17

you square both sides, then you rearrange and square both sides again it will be a quadratic.

$${you}\:{square}\:{both}\:{sides},\:{then}\:{you} \\ $$$${rearrange}\:{and}\:{square}\:{both}\:{sides}\:{again} \\ $$$${it}\:{will}\:{be}\:{a}\:{quadratic}.\: \\ $$

Commented by math solver last updated on 16/Oct/17

how did you see directly the quadratic equation?

$${how}\:{did}\:{you}\:{see}\:{directly}\:{the}\:{quadratic}\: \\ $$$${equation}? \\ $$

Commented by ajfour last updated on 16/Oct/17

if you square you′ll have to see if an obtained root is acceptable or not.

$${if}\:{you}\:{square}\:{you}'{ll}\:{have}\:{to}\:{see} \\ $$$${if}\:{an}\:{obtained}\:{root}\:{is}\:{acceptable} \\ $$$${or}\:{not}. \\ $$

Commented by math solver last updated on 16/Oct/17

exactly ! you cant directly argue for 2 solu solutions ...

$${exactly}\:!\:{you}\:{cant}\:{directly}\:{argue}\:{for}\:\mathrm{2}\:{solu} \\ $$$${solutions}\:... \\ $$

Commented by squidward last updated on 17/Oct/17

yes i can. it wasnt specified if the solutions have to be real. hence there are exactly 2 solutions either imaginary or not

$${yes}\:{i}\:{can}.\:{it}\:{wasnt}\:{specified}\:{if}\:{the} \\ $$$${solutions}\:{have}\:{to}\:{be}\:{real}.\:{hence}\:{there} \\ $$$${are}\:{exactly}\:\mathrm{2}\:{solutions}\:{either}\:{imaginary} \\ $$$${or}\:{not} \\ $$

Commented by math solver last updated on 17/Oct/17

you mean the answer of the above pro blem should be 2 then?

$${you}\:{mean}\:{the}\:{answer}\:{of}\:{the}\:{above}\:{pro} \\ $$$${blem}\:{should}\:{be}\:\mathrm{2}\:{then}? \\ $$

Commented by squidward last updated on 17/Oct/17

yes, first root is −(1/(11)) and second −3 (considering imaginary space). try substituting them yourself, you will see

$${yes},\:{first}\:{root}\:{is}\:−\frac{\mathrm{1}}{\mathrm{11}}\:{and}\:{second}\:−\mathrm{3} \\ $$$$\left({considering}\:{imaginary}\:{space}\right).\: \\ $$$${try}\:{substituting}\:{them}\:{yourself},\:{you} \\ $$$${will}\:{see} \\ $$

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