Solve the equation $t^{2} + 10 = 6 t$ $t^2+10=6t$ by completing square method?

Question

2022 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-27T06:00:34

$t = 3 - i$ $t=3-i$ or $t = 3 + i$ $t=3+i$

$t^{2} + 10 = 6 t$ $t^2+10=6t$ can be written as

$t^{2} - 6 t + 10 = 0$ $t^2-6t+10=0$

or $t^{2} - 6 t + 9 + 1 = 0$ $t^2-6t+9+1=0$

Now we use the identity ${(x + 1)}^{2} = x^{2} + 2 x + 1$ $(x+1)^2=x^2+2x+1$ and imaginary number $i$ $i$ defined by $i^{2} = - 1$ $i^2=-1$

or $t^{2} - 2 \times 3 t + 3^{2} - (- 1) = 0$ $t^2-2xx3t+3^2-(-1)=0$

or ${(t - 3)}^{2} - i^{2} = 0$ $(t-3)^2-i^2=0$

Now using identity $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ , this becomes

$(t - 3 + i) (t - 3 - i) = 0$ $(t-3+i)(t-3-i)=0$

Hence, either $t - 3 + i = 0$ $t-3+i=0$ i.e. $t = 3 - i$ $t=3-i$

or $t - 3 - i = 0$ $t-3-i=0$ i.e. $t = 3 + i$ $t=3+i$

Solve the equation t2+10=6tt^2+10=6t by completing square method?