If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots
Solution:
Given, the coefficient of x² and the constant term have opposite signs.
We have to determine if the equation has real roots.
Consider the standard form of a quadratic equation is ax2 + bx + c = 0.
coefficient of x² = ±a
coefficient of x = b
Constant term = ±c
Case 1) if coefficient of x² = a, constant term = -c
Discriminant = b² - 4ac
= b² - 4a(-c)
= b² + 4ac > 0
Therefore, the equation has 2 distinct real roots.
Case 2) if coefficient of x² = -a, constant term = c
Discriminant = b² - 4ac
= b² - 4(-a)c
= b² + 4ac > 0
Therefore, the equation has 2 distinct real roots.
In both cases, the discriminant is always positive.
✦ Try This: Determine the nature of the quadratic equation 2x² + x - 1 = 0.
Given, the equation is 2x² + x - 1 = 0
We have to determine if the equation has two distinct real roots.
Discriminant = b² - 4ac
Here, a = 2, b = 1 and c = -1
b² - 4ac = 1 - 4(2)(-1) = 1 + 8
= 9 > 0
Therefore, the equation has 2 distinct and real roots
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 4
NCERT Exemplar Class 10 Maths Exercise 4.2 Problem 2 (v)
If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots
Summary:
The assumption “If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots” is true
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