How to solve this irrational equation?

Question

How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$

Answer 1

I don't really recommend the following, but a general ansatz for $$\sqrt P - \sqrt Q = R$$ where $P$, $Q$, and $R$ are polynomials in a single unknown, might be to multiply with the conjugate: $$ P - Q = R\sqrt P + R\sqrt Q$$ then expand using $\sqrt P = R+\sqrt Q$ to get $$P - Q = R^2 + 2R\sqrt Q $$ and thus $$ (P - Q - R^2)^2 = 4R^2 Q$$ so the solutions you seek will be roots of some real polynomial -- but some of the roots will be non-solutions, and the final polynomial will generally have inconveniently high degree.

You can always do this and apply the rational root test to see if there are any nice solutions to the original equation, though. If not, then you might as well just solve the original numerically.

Answer 2

Write:

$$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$

As:

$$(\sqrt{8x + 1} - \sqrt{6x - 2}) = (2x^2 - 8x + 7).$$

Squaring both sides yields:

$$14 x-2 \sqrt{6 x-2}\sqrt{8 x+1}-1 = 4x^4-32x^3+92x^2-112x+49$$

Simplifying:

$$-2 \sqrt{6 x-2}\sqrt{8 x+1} = 4x^4-32x^3+92x^2-112x+49$$

Squaring both sides again yields:

$$ 4(6x-2)(8x+1) = 16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+25076 x^2-12600 x+2500$$

Simplifying:

$$16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+24884 x^2-12560 x+2508 = 0$$

This gives us more roots, so care must be taken, but indeed, we get back two of the $8$ roots as $x = 1$ and $x = 3$.

Numerical methods would also have worked earlier and guessing too.

Regards

Answer 3

Note that, with $x=1$, $y=\sqrt{8x+1}$ takes the value $y=3$ and with $x=3$, $y=\sqrt{8x+1}$ takes the value $y=5$. Put the point $A(1, 3)$ and $B(3,5)$. The equation of the line passing two points $A$ and $B$ is $y = x + 2.$ And then we write $$\sqrt{8x + 1}-(x+2).$$ Similar to with $$(x+1- \sqrt{6x - 2}).$$ We write the given equation has the form $$ \sqrt{8x + 1}-(x+2) + (x+1- \sqrt{6x - 2}) =2(x^2 - 4x + 3). $$ equavalent to $$ \dfrac{-(x^2 -4x + 3)}{ \sqrt{8x + 1}+(x+2)}+ \dfrac{x^2 -4x + 3}{x+1 + \sqrt{6x - 2}}=2(x^2 - 4x + 3).$$ Or $$ (x^2 -4x + 3)\left(\dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 + \sqrt{6x - 2}}\right)=0.$$ With $x\geqslant \dfrac{1}{3}$, it is easy to see that $$ \dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 + \sqrt{6x - 2}} >0.$$ Therefore we get $$ x^2 -4x + 3 = 0.$$