# Given the equation tx^2+3tx-x=-3-t, find t if the equation has 2 real solutions.

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### 3 Answers

Given the equation:

tx^2 + 3tx - x = -3 -t

We will rewrite so the right side is 0.

==> tx^2 + 3tx -x +3+t = 0

==> tx^2 + (3t-1)x + ( 3+t) = 0

Now we need to find the values of t such that the equation has 2 real solutions.

We know that the equation has two real solutions if delta > 0.

delta = b^2 - 4ac > 0

a = t b= 3t-1 c = 3+ t

==> (3t-1)^2 - 4*t*(3+t) > 0

Let us expand:

==> (9t^2 -6t + 1 -12t - 4t^2 > 0

==> 5t^2 - 18t + 1 > 0

Now we will solve for t using the formula:

==> t1= ( 18 + sqrt(324-4*5*1) / 2*5

= ( 18+ sqrt304) / 10 = (18 + 4sqrt19)/10

= (9+2sqrt19)/5

==> t2= ( 9-2sqrt19)/5

==> ( t - (9-2sqrt19)/5 ) ( t - (9+2sqrt19)/5 ) > 0

Then,

( t - (9-2sqrt19)/5 > 0 and t-(9+2sqrt19)/5 > 0

==> t > (9+2sqrt19)/5............(1)

( t-( 9-2sqrt5)/19 < 0 and t-(9+2sqrt19/5) < 0

==> t < (9-2sqrt5)/5 < 0................(2)

Then, from (1) and (2) we conclude that t values belongs to the following interval:

**==> t= ( -inf, (9-2sqrt19)/5 ) U ( 9+2sqrt19)/5 , inf)**

**OR:**

**==> t = R - [ (9-2sqrt19)/5 , ( 9+2sqrt19)/5 ]**

For a quadratic equation ax^2 + bx + c = 0, if the equation has 2 real roots D = b^2 - 4ac >0.

Here, we have tx^2 + 3tx - x = -3 -t

=> tx^2 + 3tx - x +3 +t = 0

=> tx^2 + x ( 3t - 1) + 3 +t

So b = 3t - 1, a = t and c = 3 + t

b^2 - 4ac = ( 3t - 1)^2 - 4*t*(3 + t)

=> 9t^2 + 1 - 6t - 12t - 4t^2

=> 5t^2 - 18t + 1 >0

The roots of 5t^2 - 18t + 1 = 0 are

t1 = [18 + sqrt (324 - 20)] / 10

=> t1 = (18 + 4 sqrt 19)/ 10

=> t1 = 1.8 + (sqrt 19)/5

t2 = 1.8 - (sqrt 19)/5

But as we are looking for 5t^2 - 18t + 1 >0, we have all the values from - infinite to 1.8 - (sqrt 19)/5 and 1.8 + (sqrt 19)/5 to infinite.

**Therefore t can take all values except those in (1.8 - (sqrt 19)/5, 1.8 + (sqrt 19)/5)**

We'll move all terms to the left side and we'll factorize by x:

tx^2 + x(3t-1) + 3 + t = 0

For the given equation to have real solutions, the discriminant delta has to be positive or zero.

delta = (3t-1)^2 - 4t(t+3)

delta >= 0

We'll expand the square and we'll remove the brackets in the expression of delta:

9t^2 - 6t + 1 - 4t^2 - 12t >= 0

We'll combine like terms:

5t^2 - 18t + 1 >= 0

We'll have to determine the roots of the expression 5t^2 - 18t + 1:

5t^2 - 18t + 1 = 0

We'll apply the quadratic formula:

t1 = [18+sqrt(324 - 20)]/10

t1 = (18+4sqrt19)/10

t1 = (9+2sqrt19)/5

t2 = (9-2sqrt19)/5

**The expression of delta is positive, 5t^2 - 18t + 1 >= 0, for the values of t located in the intervals:**

**(-infinite ; (9-2sqrt19)/5] U [(9+2sqrt19)/5 ; +infinite)**