Question

if a + 1 upon a equals to 17 upon 4 find the value of a minus one upon a​

Answer 1

$a+\frac{1}{a} = \frac{17}{4} \\ \\ 4a^2 + 4 = 17a \\ \\ 4a^2-17a+4=0\\\\4a^2-16a-a+4=0\\\\4a (a-4)-1(a-4)=0\\\\(a-4)(4a-1)=0\\\\a=4 \\ \\ a- \frac{1}{a} = 4 - \frac{1}{4} \\\\\frac {16-1}{4}\\\\\frac {15}{4}\\\\for\:a=\frac {1}{4}\\\\the\:value\:is\:-\frac{15}{4}$

hope it helps.

Answer 2

Given :

$a+\dfrac{1}{a}=\dfrac{17}{4}$

To find :

$a-\dfrac{1}{a}=?$

Solution :

$a+\dfrac{1}{a}=\dfrac{17}{4}$

on squaring both sides , we get :

$a^2+\dfrac{1}{a^2}=(\dfrac{17}{4})^2-2\\\\a^2+\dfrac{1}{a^2}=\dfrac{289}{16}-2\\\\a^2+\dfrac{1}{a^2}=\dfrac{289-32}{16}=\dfrac{257}{16}$

Now, we subtract 2 on both the sides, we get that

$a^2+\dfrac{1}{a^2}-2=\dfrac{257}{16}-2\\\\(a-\dfrac{1}{a})^2=\dfrac{257-32}{16}\\\\(a-\dfrac{1}{a})^2=\dfrac{225}{16}\\\\(a-\dfrac{1}{a})=\sqrt{\dfrac{225}{16}}\\\\a-\dfrac{1}{a}=\dfrac{15}{4}$

Hence, the value of $a-\dfrac{1}{a}=\dfrac{15}{4}$