Question

tsquare -t-4 find the value of t by using sridharacharya formula​

Answer 1

Step-by-step explanation:

here ,

$given \: \: equtioin \: is \: \\ \\ t {}^{2} - t - 4 = 0...............(1) \\ \\ equation \:(1) \: \: compare \: with \\ \\ ax {}^{2} + bx + c = 0 \\ \\ so \: \: a = 1 \\ \: \: \: \: \: \: b = - 1 \\ \: \: \: \: \: \: c = - 4$

sidraracharya formula is ,

x = -b+squareroot(4ac) / 2 (a)

or

x = -b--squareroot(4ac) / 2 (a)

where

alpha = b^2 -- 4ac

alpha = 1 + 16

alpha = 17

so from formula

x = -(-1)+squareroot 17 / 2 (1)

or

x = -(-1)--squareroot 17 / 2 (1)

so answer is....

x = 1+squareroot 17 / 2

or

x = 1- squareroot 17 / 2