MHB CTS and show the roots in this form

[ai93]

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$$\sqrt{5}$$

 

[evinda]

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$$\sqrt{5}$$

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

 

[soroban]

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]

 

[ai93]

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

Thanks this really helped!

 

[ai93]

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!