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Squaring on both sides we get the equation $$2x^2+7x-4=0$$ Then $$x=\frac 12$$ and $$x=-4$$ They do indeed satisfy the original equation but aren’t a part of the answer. I get that it must have to be the $\pm \sqrt {a}$ but I would still like a thorough explanation

Thanks!

Aditya
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  • We note that both solutions does not satisfy the original equation. This is because when you square both sides, you lose information on the sign of the radical since $(\sqrt{a})^2=(-\sqrt{a})^2$ – David Nov 17 '19 at 04:51
  • @David But doesn’t the square root give two values ie. $\pm$? It does satisfy if we use that logic – Aditya Nov 17 '19 at 04:56
  • any solution of the given equation would have to satisfy the quadratic you find, but the solutions of your quadratic give a negative on the RHS, which won’t be equal to the implied positive branch of the square root on the LHS, so there is no solution. – Robo300 Nov 17 '19 at 04:56
  • We have to say plus or minus because the square root function can only pick one value to be called a function, so it’s defined to pick positive. – Robo300 Nov 17 '19 at 04:58
  • A way to check is to graphically view the intersection of the functions on either side of the equal sign. – Andrew Chin Nov 17 '19 at 05:26

3 Answers3

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They do not in fact satisfy the original equation.

For $x=\frac{1}{2}$, you get $\frac{5}{2}=-\frac{5}{2}$ and $x=4$, you get $7=-7$. Both are clearly false.

A common misconception is assuming that $\sqrt{a}$ assumes two values, $\pm\sqrt{a}$ simultaneously because it is the 'inverse' of the square function. However, this is not true, and we only take the positive branch of the square root function. This is because we want a function to be well-defined, to map to a unique number, instead of 2.

David
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Hint

As $x-3=\sqrt{3x^2+x+5}\ge0$ for real $x$

$x\ge3$

lab bhattacharjee
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Solving the equation $$2x^2+7x-4=0$$ we get $$x^2+\frac{7}{2}x-2=0$$ (dividing by $2$): and by the quadratic formula $$x_{1,2}=-\frac{7}{4}\pm\sqrt{\frac{49}{16}+2}$$ $$x_{1,2}=-\frac{7}{4}\pm\frac{9}{4}$$ we get $$x_1=\frac{1}{2}$$ or $$x_2=-4$$ but for the squaring we need the condition $$x\geq 3$$ so we get no solution.

Dr. Sonnhard Graubner
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