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Suppose a quadratic equation has the form x^2 + x + c = 0. Show that the constant c must be less than 1/4 in order for the equation to have two real solutions.

Suppose A Quadratic Equation Has The Form X2 X C 0 Show That The Constant C Must Be Less Than 14 In Order For The Equation To Have Two Real Solutions class=

Sagot :

Answer:

see explanation

Step-by-step explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

Then for the equation to have 2 real roots , the discriminant must be greater than zero , that is

b² - 4ac > 0

x² + x + c = 0 ← is in standard form

with a = 1, b = 1, c = c , then

b² - 4ac > 0

1² - (4 × 1 × c) > 0

1 - 4c > 0 ( subtract 1 from both sides )

- 4c > - 1

Divide both sides by - 4, reversing the inequality as a result of dividing by a negative quantity.

c < [tex]\frac{1}{4}[/tex]