Can we solve $\sqrt{4x+13}-\sqrt{x+1}=\sqrt{12-x}?$

Can we solve $\sqrt{4x+13}-\sqrt{x+1}=\sqrt{12-x}?$

Question

1. Solve $\sqrt{4x+13}-\sqrt{x+1}=\sqrt{12-x}.$

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1. One root of the equation $3x^3+14x^2+2x-4=0$ is rational. Obtain this root and complete the solution of the equation.

Suggestion

How could we get rid of the square roots?

Squaring both sides of this will still leave a square root, but could we rearrange and square again?

If we do this, will our solutions all be valid?

What does the question mean by ‘rational’?

We could write this root as $\dfrac{a}{b}$, where the fraction is in lowest terms. What happens if we substitute this into the cubic and simplify?

Since $a$ and $b$ have no common factors, what does this tell us?

Can you find any ways to minimise the number of possibilities for $a$ and $b$ that need checking? Can we then check all the possibilities?

Solution

Squaring both sides of the equation gives us

$$\begin{align*} &(4x+13)+(x+1)-2\sqrt{(4x+13)(x+1)} = 12-x\\ \Longrightarrow\quad& 5x+14-12+x = 2\sqrt{(4x+13)(x+1)}\\ \Longrightarrow\quad& 3x+1 = \sqrt{(4x+13)(x+1)}. \end{align*}$$

Squaring again implies that

$$\begin{align*} &9x^2+6x+1 = (4x+13)(x+1)=4x^2+17x+13 \\ \Longrightarrow\quad& 5x^2-11x-12 = 0 \\ \Longrightarrow\quad& (5x+4)(x-3) = 0 \end{align*}$$

and so $x = -\dfrac{4}{5}$ or $x = 3$.

We know now that these are the only two possibilities for a solution. There is a problem, however; squaring may have intrduced false roots.

For example, if we have $x=\sqrt{x+2}$, then squaring gives $x^2 = x+2$, and so $(x-2)(x+1)=0$, and $x = 2$ or $-1$.

But $2$ works in our original equation, while $-1$ does not. So $-1$ is a false root, and $2$ is the only true solution.

So we need to check whether or not the possible solutions above satisfy our original equation.

If $x = -\dfrac{4}{5}$, then the left-hand side is $\sqrt{4\left(-\dfrac{4}{5}\right)+13}-\sqrt{-\dfrac{4}{5} + 1} = \sqrt{\dfrac{49}{5}}-\sqrt{\dfrac{1}{5}}=\dfrac{1}{\sqrt{5}}(7-1)=\dfrac{6}{\sqrt{5}},$ while the right-hand side is $\sqrt{12+\dfrac{4}{5}}=\sqrt{\dfrac{64}{5}}=\dfrac{8}{\sqrt{5}},$ which are not equal. Thus $x=-\dfrac{4}{5}$ is not a solution.

If $x=3$, then the left-hand side is $\sqrt{4\times 3+13}-\sqrt{3 + 1}=5-2=3$ while the right-hand side is $\sqrt{12-3}=\sqrt{9}=3.$ Thus $x=3$ is the only solution to the original equation.

1. One root of the equation $3x^3+14x^2+2x-4=0$ is rational. Obtain this root and complete the solution of the equation.

Let $f(x)=3x^3+14x^2+2x-4$. Assume the given rational root is $\dfrac{a}{b}$, where the highest common factor of $a$ and $b$ is $1$ and, without loss of generality, $b>0$.

Substituting this solution into $f(x)=0$ and multiplying through by $b^3$ we find $3a^3+14a^2b+2ab^2-4b^3=0.$ Note that this means $b$ divides $3a^3$ (since it divides every other term) and so $b$ is $1$ or $3$ (since it cannot share a factor with $a$).

Now $2$ divides $14a^2b+2ab^2-4b^3$, and so $2$ divides $3a^3$, and thus $2$ divides $a$, say $a = 2c$. This gives us $6c^3 + 14c^2b + cb^3 - b^3 = 0,$ and so $c$ divides $b^3$. But $c$ and $b$ have no common factor, and so $c$ must be $\pm 1$, and $a$ must be $\pm 2$.

So the only possible rational roots for the equation are $\dfrac{2}{3}, -\dfrac{2}{3}, 2$ or $-2$.

It is easy to check that $-\dfrac{2}{3}$ is the only one to satisfy the equation. Thus $x=-\dfrac{2}{3}$ is the rational root. Factorising the cubic we find $3x^3 + 14x^2 + 2x - 4 = (3x+2)(x^2+4x-2)=0.$ So either $x = -\dfrac{2}{3}$, or $x^2+4x-2=0$ and so $x=\dfrac{-4\pm\sqrt{16-4\times(-2)}}{2}=-2\pm\sqrt{6}.$