If the coefficient of x² and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots
Solution:
Given, the coefficient of x² and the constant term have opposite signs.
We have to determine if the equation has no real roots.
Consider the standard form of a quadratic equation is ax2 + bx + c = 0.
coefficient of x² = ±a
coefficient of x = 0
Constant term = ±c
Case 1) if coefficient of x² = -a, constant term = -c
Discriminant = b² - 4ac
= 0 - 4(-a)(-c)
= -4ac > 0
Therefore, the equation has no real roots.
Case 2) if coefficient of x² = a, constant term = c
Discriminant = b² - 4ac
= 0 - 4ac
= -4ac < 0
In both cases, the discriminant is always negative.
Therefore, the equation has no real roots.
✦ Try This: Determine the roots of the equation 3x² + 1 = 0.
Given, the equation is 3x² + 1 = 0
We have to determine if the equation has two distinct real roots.
Discriminant = b² - 4ac
Here, a = 3, b = 0 and c = 1
b² - 4ac = 0 - 4(3)(1)
= -12 < 0
We know that a quadratic equation ax² + bx + c = 0 has no real roots when the discriminant of the equation is equal to zero.
Therefore, the equation has no real roots
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 4
NCERT Exemplar Class 10 Maths Exercise 4.2 Problem 2 (vi)
If the coefficient of x² and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots
Summary:
The assumption “if the coefficient of x² and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots” is true
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