F(x)

Find $x$ :

$\large \sqrt[3]{x - 4} +\sqrt{x-3}=5$

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`2 \times 3`	$2 \times 3$
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Comments

$\sqrt[3]{x-4} = 5 - \sqrt{x-3}$ .

Cube both sides:

$x-4 = \left(5 - \sqrt{x-3} \right)^2 = 25 + (x-3) - 10 \sqrt{x-3}$

$-26 = -10 \sqrt{x-3}$

Can you finish it from here?

Pi Han Goh - 3 years, 6 months ago

@Pi Han Goh , you have to cube , not square. Please correct your solution.

Vilakshan Gupta - 3 years, 6 months ago

Let $\sqrt[3]{x-4} = k$ , therefore $x-3=k^3+1$ and new equation becomes $\begin{aligned} k+\sqrt{k^3+1} = 5 \\ & \implies \sqrt{k^3+1}=5-k \\ &\implies k^3+1=k^2-10k+25 &&&&& \color{#3D99F6}\text{(squaring both sides)} \\ &\implies k^3-k^2+10k-24=0 \\ &\implies (k-2)(k^2+k+12)=0 &&&&& \color{#3D99F6}\text{(note that k^2+k+12 > 0)} \\ &\implies k=2 \implies x=\boxed{12} \end{aligned}$

Vilakshan Gupta - 3 years, 6 months ago

@Vilakshan Gupta – Whoops. Haha, my bad. Thanks for fixing it.

Pi Han Goh - 3 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$