Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$

Question

Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$

---

I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?

Answer 1

Let $z:=\sqrt{3x-4}\ge0$. We have

$$z+\sqrt[3]{1-z^2}=1$$

or by cubing,

$$(z-1)^3+(1-z^2)=z^3-4z^2+3z=z(z-1)(z-3)=0.$$

This yields three solutions in $x$,

$$\frac43,\frac53,\frac{13}3.$$

Answer 2

$$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$$ $$ \sqrt[3]{5 - 3x} = 1-\sqrt{3x - 4}$$ $$ 5 - 3x = (1-\sqrt{3x - 4})^3$$ $$ 5 - 3x = (1-3x)\sqrt{3x-4}+9x-11$$ $$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$ $$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$ $$ -4\sqrt{3x-4}\cdot \sqrt{3x-4}= (1-3x)\sqrt{3x-4}$$

so $x=4/3$ or: $$ -4 \sqrt{3x-4}=1-3x$$

which is easier. You can finish?

Answer 3

By $x=\frac {y^3} 3+\frac 53$

$$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1 \iff \sqrt {y^3+1}-y=1 $$

$$\iff y^3+1=(1+y)^2 \iff y^3-y^2-2y=0 \iff y\in \{-1,0,2\}$$

that is $x\in \{\frac43,\frac53,\frac{13}3\}$.