CTS and show the roots in this form

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Discussion Overview

The discussion revolves around solving the quadratic equation $$x^{2}-8x-29=0$$ and expressing its roots in the form of $$c \pm d\sqrt{5}$$. Participants explore the method of completing the square (CTS) to derive the roots, focusing on the steps involved in the transformation of the equation.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • Some participants describe using the completing the square method, arriving at the equation $$(x-4)^2-45=0$$ as a step in their solution.
  • Participants express uncertainty about the next steps to express the roots in the desired form of $$c \pm d\sqrt{5}$$ after reaching the equation $$(x-4)^2=45$$.
  • One participant clarifies that solving for $$x$$ involves taking the square root of both sides, leading to $$x-4=\pm\sqrt{45}$$ and subsequently to $$x=4 \pm 3\sqrt{5}$$.
  • Another participant thanks others for their contributions, indicating that the explanations helped clarify the process.

Areas of Agreement / Disagreement

Participants generally agree on the method of completing the square and the steps to derive the roots, but there is some repetition and uncertainty about the clarity of the process, particularly in expressing the roots in the specified form.

Contextual Notes

Some steps in the mathematical reasoning are reiterated across multiple posts, and there is a focus on the purpose of the completing the square method without resolving any potential confusion about its application.

ai93
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I have to show the roots of $$x^{2}-8x-29=0$$ are c$$\pm$$d$$\sqrt{5}$$

I used completing the square method. Once I used CTS I got the answer
$$(x-4)^2-45=0$$ So I am not sure what is the next step to put it in the form of c$$\pm$$d$$\sqrt{5}$$
 
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mathsheadache said:
I have to show the roots of $$x^{2}-8x-29=0$$ are c$$\pm$$d$$\sqrt{5}$$

I used completing the square method. Once I used CTS I got the answer
$$(x-4)^2-45=0$$ So I am not sure what is the next step to put it in the form of c$$\pm$$d$$\sqrt{5}$$

$$(x-4)^2=45 \Rightarrow x-4= \pm \sqrt{45} \Rightarrow x-4=\pm \sqrt{9 \cdot 5} \Rightarrow x-4=\pm 3 \sqrt{5} \Rightarrow x=4 \pm 3 \sqrt{5}$$
 
Hello, mathsheadache!

I have to show the roots of: $$\:x^2-8x-29\:=\:0\:$$ are [math]\:c\pm d\sqrt{5}[/math]

I used completing the square method.
Once I used CTS I got the answer: $$\;(x-4)^2-45\:=\:0$$

So I am not sure what is the next step to put it in the form of [math]c\pm d \sqrt{5}[/math]
What is the purpose of CTS? . . . To solve for [math]x\,![/math]

[math](x-4)^2 - 45 \;=\;0[/math]

[math]\qquad(x-4)^2 \;=\; 45[/math]

[math]\qquad\quad x-4 \;=\;\pm\sqrt{45}[/math]

[math]\qquad\quad x-4 \;=\;\pm3 \sqrt{5}[/math]

[math]\qquad\qquad\;\, x \;=\;4 \pm 3\sqrt{5}[/math]
 
evinda said:
$$(x-4)^2=45 \Rightarrow x-4= \pm \sqrt{45} \Rightarrow x-4=\pm \sqrt{9 \cdot 5} \Rightarrow x-4=\pm 3 \sqrt{5} \Rightarrow x=4 \pm 3 \sqrt{5}$$

Thanks this really helped!
 
soroban said:
Hello, mathsheadache!


What is the purpose of CTS? . . . To solve for [math]x\,![/math]

[math](x-4)^2 - 45 \;=\;0[/math]

[math]\qquad(x-4)^2 \;=\; 45[/math]

[math]\qquad\quad x-4 \;=\;\pm\sqrt{45}[/math]

[math]\qquad\quad x-4 \;=\;\pm3 \sqrt{5}[/math]

[math]\qquad\qquad\;\, x \;=\;4 \pm 3\sqrt{5}[/math]

Cheers, you make the question look easier than it is!
 

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