Question: "(t+2)23+3(t+2)13−10=0 how do i solve?"

High School
Algebra
Algebra I
Quadratic function and equation

Lipossig

Answered question

2021-06-07

\frac{{(t + 2)}^{2}}{3} + 3 \frac{{(t + 2)}^{1}}{3} - 10 = 0

how do i solve?

Answer & Explanation

BleabyinfibiaG

Skilled2021-06-08Added 118 answers

Given equation is: $\frac{{(t + 2)}^{2}}{3} + 3 (t + 2) \frac{1}{3} - 10 = 0 \dots (1)$
Let $\frac{{(t + 2)}^{1}}{3} = x$
$\frac{{(t + 2)}^{1}}{3} (t + 2) 13 + 3 (t + 2) 13 - 10 = 0$
$x \times x + 3 \times x - 10 = 0$
$x^{2} + 3 x - 10 = 0 \dots (2)$
Solving the quadratic equation (2) $x^{2} + 3 x - 10 = 0$
$x^{2} + 5 x - 2 x - 10$
$x (x + 5) - 2 (x + 5) = 0$
$(x + 5) (x - 2) = 0$
$x = - 5$ and $x = 2$
Putting the value of x : $\frac{{(t + 2)}^{1}}{3} = x$
$\frac{{(t + 2)}^{1}}{3} = 2$
cubing both the sides $(t + 2) = 2^{3}$
$(t + 2) = 8$
$t = 8 - 2 = 6$
$t = 6$
$\frac{(t + 2)^{1}}{3} = - 5$
$(t + 2) = - 125$
$t = - 125 - 2 = - 127$
Thus the value of t are 6 and -127

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra I

Didn't find what you were looking for?