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Question Number 217842 by aylinrabbani last updated on 22/Mar/25

$$f\left(x\right)=\sqrt{x^2-6x+10}$$$$f(3+2\sqrt{6})=?$$

Answered by Rasheed.Sindhi last updated on 22/Mar/25

$$f\left(x\right)=\sqrt{x^2-6x+10}$$$$=\sqrt{x^2-6x+9+1}$$$$=\sqrt{{(x-3)}^2+1}$$$$f(3+2\sqrt{6})=\sqrt{{(3+2\sqrt{6}-3)}^2+1}$$$$=\sqrt{{\left(2\sqrt{6}\right)}^2+1}$$$$=\sqrt{4\left(6\right)+1}$$$$=\sqrt{25}$$$$=5$$

Answered by AntonCWX8 last updated on 22/Mar/25

$${(3+2\sqrt{6})}^2-6(3+2\sqrt{6})+10$$$$=9+12\sqrt{6}+24-18-12\sqrt{6}+10$$$$=25$$$$$$$$\sqrt{25}=\pm 5$$

Commented by mr W last updated on 22/Mar/25

$$\sqrt{25}=5\ne \pm 5$$

Commented by Frix last updated on 22/Mar/25

$$\mathrm{If}\sqrt{25}=\pm 5\mathrm{then}\sqrt{25}+\sqrt{25}=?$$$$\left(\mathrm{a}\right)-10$$$$\left(\mathrm{b}\right)0$$$$\left(\mathrm{c}\right)10$$

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