Exam question

Q) Using the curve $ y = 2x^3 - 3x +2 $ find the solutions to the equation $4x^3 - 6x +1 = 0 $ [3 marks]

The answer given was as follows:

$2x^3 - 3x +2 = -2x^3 +3x +1 \implies \text{ I can understand this step but not the next } \\ 2x^3 - 3x +2 = 1.5 \implies \text{ from where do you get 1.5 }$

After that, they just drew $y = 1.5$ and the intersection points gave the solution but my question is how do you get $1.5$ ?

Please explain.

#Algebra

Easy Math Editor

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Comments

I would try to solve it in the following way. $4x^3-6x+1=0$ $\implies 4x^3-6x+4=3$ $\implies 2y=3$ $\implies y=\frac{3}{2}$ So, $4x^3-6x+1$ becomes zero when $y = 2x^3-3x+2 = \frac{3}{2}$ . One way to find the corresponding value of $x$ is to draw $2x^3-3x+2$ and find when it intersects $y=\frac{3}{2}$ .

Atomsky Jahid - 1 year, 5 months ago

Thanks alot. It was very helpful

Syed Hamza Khalid - 1 year, 5 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}`	$\boxed{123}$