If b = 0, c < 0, is it true that the roots of x² + bx + c = 0 are numerically equal and opposite in sign
Solution:
Given, the equation is x² + bx + c = 0.
b = 0 and c < 0
We have to determine if the roots of the equation are numerically equal and opposite in sign.
Using the quadratic formula,
x = [-b ± √b² - 4ac]/2a
Here, a = 1, b = 0
Since c < 0 take c as -c.
x = [-0 ± √0 - 4(1)(-c)]/2(1)
x = ± √4c/2
x = ± c
Therefore, the roots are c and -c.
✦ Try This: Consider the equation x² - 4 = 0
Given, the equation is x² - 4 = 0
We have to determine the roots of the equation.
x² = 4
Taking square root,
x = ±2
Therefore, the roots are 2 and -2
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 4
NCERT Exemplar Class 10 Maths Exercise 4.2 Problem 7
If b = 0, c < 0, is it true that the roots of x² + bx + c = 0 are numerically equal and opposite in sign
Summary:
If b = 0, c < 0, the roots of x² + bx + c = 0 are numerically equal and opposite in sign is true.
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