How do you solve $t - \sqrt{6 t - 9} = 0$ $t - sqrt (6t - 9) = 0$ ?

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Answer 1 · 2016-02-05T13:58:01

$t = 3$ $t= 3$

Given: $t - \sqrt{6 t - 9} = 0$ $t-sqrt(6t-9)=0$ .......................(1)

Things would be easier if we 'got rid' of the square root!

$t = \sqrt{6 t - 9}$ $t=sqrt(6t-9)$

Square both sides

$t^{2} = 6 t - 9$ $t^2=6t-9$
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Make into a quadratic equation

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

Observe that $- 3 - 3 = - 6 and that (- 3) \times (- 3) = + 9$ $" " -3-3=-6 " and that " (-3)xx(-3)=+9$

Factorising gives:

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

so $t = + 3$ $" " t= +3$
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Check if correct!

Substituting for t in equation 1 and consider the left hand side only

$3 - \sqrt{6 (3) - 9}$ $3-sqrt(6(3)-9)$

$3 - \sqrt{18 - 9}$ $3-sqrt(18-9)$

$3 - (\pm 3)$ $3-(+-3)$

The only possible value for the LHS to be zero is to have:

$L H S \to 3 - 3 = 0 and R H S \to 0$ $LHS->3-3=0" and " RHS->0$ so proven to be correct

How do you solve t−√6t−9=0t - sqrt (6t - 9) = 0?