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

User Jutikorn
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1 Answer

4 votes

Answer:

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

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.

User Lucas Crawford
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