Roots

**james_bond** · November 16th 2009, 10:29 AM

Show that $(4-x)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4$ if $a=1+\sqrt5$ .

**Opalg** · November 17th 2009, 12:22 AM

Originally Posted by james_bond

Show that $(4-{\color{red}a})\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4$ if $a=1+\sqrt5$ .

If $a=1+\sqrt5$ then $a^2 = 6+2\sqrt5$ and $3a+4 = 7+3\sqrt5$ . Multiply these together to check that $a^2(3a+4) = 72+32\sqrt5$ .

Next, $(2+a)^3 = (3+\sqrt5)^3$ . check that is also equal to $72+32\sqrt5$ . Therefore $\sqrt{2+a} = \bigl(a^2(3a+4)\bigr)^{1/6} = \sqrt[3]{a}\cdot\sqrt[6]{3a+4}$ .

Therefore $(4-a)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4} = (4-a)(2+a) = (3-\sqrt5)(3+\sqrt5) = 9-5=4$ .

Thread: Roots

LinkBack

Thread Tools

Display

Roots

Similar Math Help Forum Discussions

roots

Roots & Imaginary Roots

roots

Roots ><

Given 2 imaginary roots find other 2 roots

Search Tags