Question

if M is equal to root 17 minus 4 then find the value for M square plus 1 by M
square
​

Answer 1

Answer:

= 66

Step-by-step explanation:

$M=\sqrt{17}-4 \\\\ M^2+(\frac{1}{M})^2=? \\\\ => \frac{1}{M}=\frac{1}{\sqrt{17}-4} \\\\ rationalising \ factor=\sqrt{17}+4\\\\ => \frac{1}{M}=\frac{1}{\sqrt{17}-4} \times \frac{\sqrt{17}+4}{\sqrt{17}+4} \\\\ => \frac{1}{M}=\frac{\sqrt{17}+4}{(\sqrt{17}-4)(\sqrt{17}+4)}\\\\ => \frac{1}{M}=\frac{\sqrt{17}+4}{\sqrt{17}^2-4^2}\\\\ => \frac{1}{M}=\frac{\sqrt{17}+4}{17-16} \\\\ =>\frac{1}{M}=\frac{\sqrt{17}+4}{1} \\\\ =>\frac{1}{M}=\sqrt{17}+4$

$\implies M^2+(\frac{1}{M})^2=[\sqrt{17}-4]^2+[\sqrt{17}+4}]^2 \\\\ \implies M^2+(\frac{1}{M})^2=\sqrt{17}^2 +4^2-2(\sqrt{17})(4) +\sqrt{17}^2 +4^2+2(\sqrt{17})(4) \\\\ \implies M^2+(\frac{1}{M})^2=17+16+17+16 \\\\ \implies M^2+(\frac{1}{M})^2=66$

$FORMULA \ USED: \\\\ (a+b)(a-b)=a^2-b^2 \\ (a+b)^2=a^2+b^2+2ab \\ (a-b)^2=a^2+b^2-2ab$

hope it helps