Tx²+3x-7=0 has two real solution. what can be the deducted about the value of t?

value of t is greater than equal to -9 / 28.

Step-by-step explanation:

Given Quadratic Polynomial :  tx² + 3x - 7 = 0

Also, It has real solutions.

Standard Quadratic equation, is ax² + bx + c = 0

here, Determinant, D = b² - 4ac

decides nature of the roots.

if D < 0 , roots / solutions are complex

if D = 0 , roots are real and equal.

if D > 0 , roots are real and different.

As given roots are real solutions.

Means Dis either equal to 0 or greater than 0

when D = 0

we have, 3² - 4 × t × (-7) = 0

               9 + 28t = 0

               t = -9 / 28

when D > 0

we have, 3² - 4 × t × (-7) > 0

               9 + 28t > 0

               t > -9 / 28

Therefore, value of t is greater than equal to -9 / 28.

center= (-2,0)

radius= sqrt2/2

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.