(x - √2)² - 2(x + 1) = 0. State whether the following quadratic equation has two distinct real roots
Solution:
Given, the equation is (x - √2)² - 2(x + 1) = 0
We have to determine if the equation has two distinct real roots.
By using algebraic identity,
(a - b)² = a² - 2ab + b²
(x - √2)² = x² - 2√2x + 2
2(x + 1) = 2x + 2
So, x² - 2√2x + 2 -(2x + 2) = 0
x² - 2√2x + 2 - 2x - 2 = 0
x² - 2√2x - 2x = 0
x² - (2√2 + 2)x = 0
Discriminant = b² - 4ac
Here, a = 1, b = -(2√2 + 2) and c = 0
b² - 4ac = [-(2√2 + 2)]² - 4(1)(0)
= (2√2 + 2)²
= 2²(√2 + 1)²
= 4(2.414)²
= 23.31 > 0
We know that a quadratic equation ax² + bx + c = 0 has 2 distinct real roots when the discriminant of the equation is greater than zero.
Therefore, the equation has 2 distinct real roots.
✦ Try This: Determine the nature of the quadratic equation 7x² + 2x - 3 = 0
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 4
NCERT Exemplar Class 10 Maths Exercise 4.2 Problem 1 (vi)
(x - √2)² - 2(x + 1) = 0. State whether the following quadratic equation has two distinct real roots
Summary:
The equation (x - √2)² - 2(x + 1) has 2 distinct real roots since the discriminant of the equation is greater than zero.
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