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30 November, 18:50

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

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  1. 30 November, 21:06
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    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.
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