Where is the mistake?

Given that pp is a real number , then for what values of pp the equation (x+3p+1)13x13=1 {(x+3p+1)}^{\frac{1}{3}} - x^ \frac{1}{3} = 1 has real solutions.

My answer:

(x+3p+1)13x131=0 {(x+3p+1)}^{\frac{1}{3}} - x^ \frac{1}{3} - 1 = 0 if a+b+c=0 a+b+c = 0 we have a3+b3+c3=3abca^3+b^3+c^3 = 3abc x+3p+1x1=3[(x+3p+1)(x)]13 x+ 3p + 1 -x -1 = 3[(x+3p+1)(x)]^\frac{1}{3} 3p=3[(x+3p+1)(x)]13 3p = 3[(x+3p+1)(x)]^\frac{1}{3} p3=x(x+3p+1) p^3 = x(x + 3p +1) x2+(3p+1)xp3=0 x^2 + (3p + 1)x -p^3 =0 For real roots D0 D \geq 0 4p3+9p2+6p+10 4p^3 + 9p^2 + 6p +1 \geq 0 (p+1)2(4p+1)0 (p+1)^2(4p+1) \geq 0 therefore , p141. p \geq \frac{-1}{4} \cup -1.

Please help me understand which step did I do wrong? And what is the correct solution for the above question?

Note by Ujjwal Mani Tripathi
3 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Now, note that

a+b+c=0    a3+b3+c3=3abc a+b+c=0 \implies a^3 + b^3 + c^3 = 3abc

But, a3+b3+c3=3abc⇏a+b+c=0 a^3 + b^3 + c^3 = 3abc \not \Rightarrow a+b+c=0

So, a+b+c=0a+b+c=0 is not equivalent to a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc.

A better approach to solve this is putting x=y3    x13=yx=y^3 \iff x^{\frac{1}{3}} = y.

Then, the equation becomes:

(y3+3p+1)13=1+y    y3+3p+1=(1+y)3    y2+yp=0\begin{aligned}(y^3 + 3p +1)^{\frac{1}{3}} &= 1+y \\\iff y^3 + 3p +1 &= (1+y)^3 \\\iff y^2 + y -p &= 0 \end{aligned}

So, for the roots of the original equation to be real, we need the discriminant of y2+yp=0y^2 + y - p=0 to be non negative.

That is, 1+4p0    p14 1+4p \geq 0 \\\iff p \geq -\frac{1}{4}

Deeparaj Bhat - 3 years, 4 months ago

Log in to reply

I got what you meant here but still i think the answer is wrong , check on wolframalpha putting p = -0.25 , it says no solution exists to the below equation for (x+0.25)13x131=0 {(x+0.25)}^{\frac{1}{3}} - x^ \frac{1}{3} - 1 = 0

Ujjwal Mani Tripathi - 3 years, 4 months ago

Log in to reply

Try x=18x = \frac{-1}{8}

Deeparaj Bhat - 3 years, 4 months ago

Log in to reply

Deeparaj Bhat - 3 years, 4 months ago

Log in to reply

@Deeparaj Bhat Oh yes , i was working only with principal root value and hence was getting no solution

Ujjwal Mani Tripathi - 3 years, 4 months ago

Log in to reply

WolframAlpha gives you the option to "use the real-valued root instead", (in blue just below the entry box), which in this case will give you a value of x=18x = -\frac{1}{8} as Deeparaj has pointed out.

Brian Charlesworth - 3 years, 4 months ago

Log in to reply

What does 141\frac{-1}{4} \cup 1 mean?

Agnishom Chattopadhyay Staff - 3 years, 4 months ago

Log in to reply

i mean that p>= -0.25 and also p = -1 is a solutionof the inequality
(p+1)2(4p+1)0 (p+1)^2(4p+1) \geq 0

Ujjwal Mani Tripathi - 3 years, 4 months ago

Log in to reply

@Brian Charlesworth @Jon Haussmann , your help will be valuable .

Ujjwal Mani Tripathi - 3 years, 4 months ago

Log in to reply

I used the same approach as Deeparaj, but I think we need more than just the discriminant to be non-negative; in order to avoid complex values for y=x13y = x^{\frac{1}{3}} we will require that x0x \ge 0 and thus also y0y \ge 0. So we require that

y=1+1+4p201+4p1p0y = \dfrac{-1 + \sqrt{1 + 4p}}{2} \ge 0 \Longrightarrow \sqrt{1 + 4p} \ge 1 \Longrightarrow p \ge 0,

which seems to be in agreement with results from WolframAlpha.

Brian Charlesworth - 3 years, 4 months ago

Log in to reply

Don't we define x13=(x)13x<0x^{\frac{1}{3}} = -(-x)^{\frac{1}{3}} \quad x < 0 right?

Deeparaj Bhat - 3 years, 4 months ago

Log in to reply

@Deeparaj Bhat Yes, we could choose to work with the real-valued root, in which case WolframAlpha does return a value of x=18x = -\frac{1}{8} for p=14p = -\frac{1}{4}, and no solutions if p<14p \lt -\frac{1}{4}.

(WolframAlpha can be a bit difficult when taking fractional roots of negative numbers.)

Brian Charlesworth - 3 years, 4 months ago

Log in to reply

@Brian Charlesworth this is exactly what i was confused about . Thank you for helping me out with the wolframalpha query .

Ujjwal Mani Tripathi - 3 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...