CTS and show the roots in this form

[ai93]

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

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

 

[evinda]

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

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

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

 

[soroban]

Hello, mathsheadache!

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

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

So I am not sure what is the next step to put it in the form of \(\displaystyle c\pm d \sqrt{5}\)

What is the purpose of CTS? . . . To solve for \(\displaystyle x\,!\)

\(\displaystyle (x-4)^2 - 45 \;=\;0\)

\(\displaystyle \qquad(x-4)^2 \;=\; 45\)

\(\displaystyle \qquad\quad x-4 \;=\;\pm\sqrt{45}\)

\(\displaystyle \qquad\quad x-4 \;=\;\pm3 \sqrt{5}\)

\(\displaystyle \qquad\qquad\;\, x \;=\;4 \pm 3\sqrt{5}\)

 

[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 \(\displaystyle x\,!\)

\(\displaystyle (x-4)^2 - 45 \;=\;0\)

\(\displaystyle \qquad(x-4)^2 \;=\; 45\)

\(\displaystyle \qquad\quad x-4 \;=\;\pm\sqrt{45}\)

\(\displaystyle \qquad\quad x-4 \;=\;\pm3 \sqrt{5}\)

\(\displaystyle \qquad\qquad\;\, x \;=\;4 \pm 3\sqrt{5}\)

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