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Solve equation: $\sqrt{t +9} - \sqrt{t} = 1$

I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$

I squared both sides $(\sqrt{t+9})^2 = (1)^2 (\sqrt{t})^2$

Then I got $t + 9 = 1+ t$

Can't figure it out after that point.

The answer is $16$

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    $\begingroup$ Hint: Look at how you squared the RHS, you have $(1 -\sqrt{t})^2$. $\endgroup$
    – Amzoti
    May 22, 2013 at 13:00
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    $\begingroup$ To follow up on @Amzoti's comment, $1-\sqrt{t}$ is a sum, not the product $(1)(-\sqrt{t})$. The square of a product is the product of the squares, but to find the square of a sum you must use $(a+b)^2=a^2+2ab+b^2$. $\endgroup$
    – vadim123
    May 22, 2013 at 13:22
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    $\begingroup$ Additionally, you can't 'move' things just by copying and pasting them. You should have added $\sqrt t$ to both sides; it cancels out on the left, and you get $1+\sqrt t$ on the right. $\endgroup$
    – not all wrong
    May 22, 2013 at 13:38
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    $\begingroup$ For some reason I feel like I should be using hyperbolic function substitution with this. Why am I thinking that? $\endgroup$
    – AncientSwordRage
    May 22, 2013 at 22:57
  • $\begingroup$ The formula $(a+b)^2=a^2+2ab+b^2$ comes from the distributive property, by the way. Here's a proof of it: The distributive property says that $c(a+b)=ca+cb$. Letting $c=a+b$, we have $(a+b)(a+b)=(a+b)a+(a+b)b$, or $(a+b)^2=(a^2+ba)+(ab+b^2)=a^2+2ab+b^2$. $\endgroup$
    – Akiva Weinberger
    Jul 21, 2015 at 3:16

7 Answers 7

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$$\sqrt{t +9} - \sqrt{t} = 1$$

Multiplying by $\sqrt{t +9} + \sqrt{t}$ you get

$$9=\sqrt{t +9} +\sqrt{t} $$

Now adding

$$\sqrt{t +9} + \sqrt{t} =9$$ $$\sqrt{t +9} - \sqrt{t} = 1$$

you get

$$\sqrt{t+9}=5 \Rightarrow t=25-9 $$

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  • $\begingroup$ @RossMillikan Fixed ty. $\endgroup$
    – N. S.
    May 22, 2013 at 13:47
  • $\begingroup$ Nice. Vote up! Small note: subtracting would give an easier equation: $2\sqrt{t} = 8$... $\endgroup$
    – johannesvalks
    Jul 20, 2015 at 23:35
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after squaring both sides you get:

$$t+9=1+2\sqrt{t}+t\implies4=\sqrt{t} \implies t=16$$

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    $\begingroup$ Your minus sign is wrong, actually. $\endgroup$
    – Jeppe Stig Nielsen
    May 22, 2013 at 21:22
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The solution of such a problem such a problem lies in finding a number for which the equation holds. Suppose you didn't know the answer was 16, how might you solve the problem?

Well, you have two square roots in the problem $\sqrt{t +9}$ and $\sqrt t$. Their squares differ by 9. Thus, if you can find two square numbers which differ by 9, then the larger square number will equal (t +9) and the smaller square number will equal t. The set of square numbers, as you might know, consists of the sequence (1, 4, 9, 16, 25, ...). 25 and 16 differ by 9, and 16 is the smaller. Thus, we have found 16 as the solution to this equation.

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    $\begingroup$ +1 for actually thinking about the problem intuitively rather than mechanically applying algebra rules. Not saying that one is better than the other but it certainly is refreshing. $\endgroup$
    – Thomas
    May 22, 2013 at 14:58
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    $\begingroup$ This approach gets lucky because the solution is a natural number. $\endgroup$
    – Reinstate Monica
    Jul 20, 2015 at 21:12
  • $\begingroup$ +1 for the simple intuitive thinking for simple problems. $\endgroup$
    – Shailesh
    Jul 20, 2015 at 23:48
  • $\begingroup$ @Solomonoff'sSecret No, it doesn't get lucky, because the solution is a natural number. Look again at the sequence of squares of natural numbers. The difference of two consecutive square numbers belongs to the sequence (1, 3, 5, 7, 9 ...) by inspection and more generally, ((n ^ 2) - ((n - 1) ^ 2)) = ((n ^ 2) - (n ^ 2 - 2 * n + 1) = 2 * n - 1, which is always odd, when n > -1. Thus, if the difference of two numbers differ by some odd number "ODD", you can always find two consecutive natural numbers which have squares that differ by ODD. $\endgroup$
    – Doug Spoonwood
    Jul 21, 2015 at 2:02
  • $\begingroup$ @DougSpoonwood You didn't state that "such a problem" only included cases where the squares differ by an odd number and nor did you state why such a restriction implies a natural number solution exists. I agree that with such a restriction, this approach doesn't get lucky. But with the more obvious generalization where 9 is replaced by an arbitrary natural or real number, this approach doesn't work. $\endgroup$
    – Reinstate Monica
    Jul 22, 2015 at 12:44
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Hint. Let $x=\sqrt{t}$ and $y=\sqrt{t+9}$. Then $$ y-x=1\implies y^2-x^2=y+x\implies 9=y+x $$

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From $\sqrt{t +9} - \sqrt{t} = 1$ you don't get $\sqrt{t +9} = 1 -\sqrt{t}$ but $$\sqrt{t +9} = 1 +\sqrt{t}.$$

You can solve as follows, using the algebraic identity $(a+b)^2=a^2+2ab+b^2$: $$ \begin{eqnarray*} \sqrt{t+9}-\sqrt{t} &=&1\Leftrightarrow \sqrt{t+9}=1+\sqrt{t}\tag{1} \\ \text{Square both sides of $(1)$} &\Rightarrow &\left( \sqrt{t+9}\right) ^{2}=\left( 1+\sqrt{t}\right) ^{2} \\ \text{Compute and simplify} &\Leftrightarrow &t+9=1+2\sqrt{t}+t\Leftrightarrow 9=1+2\sqrt{t} \\ \text{Simplify} &\Leftrightarrow &9-1=2\sqrt{t}\Leftrightarrow 8=2\sqrt{t} \\ \text{Simplify} &\Leftrightarrow &\frac{8}{2}=\sqrt{t}\Leftrightarrow 4= \sqrt{t}\tag{2} \\ \text{Square both sides of $(2)$} &\Rightarrow &4^{2}=\left( \sqrt{t}\right) ^{2} \\ &\Leftrightarrow &16=t.\tag{3} \end{eqnarray*} $$

Final comment. When we square both sides of an equation we get a new equation which has the same solutions of the original equation, but can have additional solutions. However in this case we got only the solution $t=16$, which is a solution of $(1)$ too.

ADDED. In your recent question solve the equation $\sqrt{3x-2}+2-x=0$, we get two solutions after squaring $$ \begin{equation*} \sqrt{3x-2}+2-x=0\Rightarrow 3x-2=x^{2}-4x+4\Leftrightarrow x\in \{1,6\} \end{equation*} $$ but only $x=6$ is a solution of $\sqrt{3x-2}+2-x=0$, as explained in this comment by Glen O.

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  • $\begingroup$ @Utveckla och förenkla You are welcome! $\endgroup$
    – Américo Tavares
    May 22, 2013 at 13:45
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An often overlooked fact is that $$ \sqrt{t^2}=\left|t\right|$$ Call me paranoid but here's how I would solve this $$\sqrt{t +9} - \sqrt{t} = 1$$ $$\sqrt{t +9} = \sqrt{t}+1$$ $$\left|t +9\right| = \left(\sqrt{t}+1\right)\left(\sqrt{t}+1\right)$$ $$\left|t +9\right| = \left|t\right|+2\sqrt{t}+1$$ Since the original equation has $\sqrt{t}$, then we know that $t\geq 0$ and we can safely remove the absolute value bars. So now we have $$t +9= t+2\sqrt{t}+1$$ $$9= 2\sqrt{t}+1$$ $$8= 2\sqrt{t}$$ $$4= \sqrt{t}$$ $$\left|t\right|=4^2$$ $$t=16$$

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Consinder the function $f(t) = \sqrt{t+9}-\sqrt{t}-1$, $t\geqslant 0$.

We have $f'(t) = \dfrac{1}{2\sqrt{t+9}} - \dfrac{1}{2\sqrt{t}} < 0$.

Therefore $f$ is a dececreasing function on the $(0, +\infty)$.

Another way, $t = 16$ is a solution of the given equation.

Therefore the given equation has only solution $t = 16.$

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