The quadratic equation, y = x² + 3x + 1 has two real roots.
How to define the nature of roots in a quadratic equation?
In a quadratic equation of the form, ax² + bx + c = 0, the nature of the roots depends upon the value of the discriminant (D), which is calculated by the formula, D = b² - 4ac.
If the value of D > 0, the equation has real and distinct roots.
If the value of D = 0, the equation has real and equal roots.
If the value of D < 0, the equation has imaginary roots.
How to solve the question
In the question, we are given an equation y = x² + 3x + 1 and are asked to say how many real roots it has.
To find the roots, we first equate y to 0.
∴ y = x² + 3x + 1 = 0.
Now comparing the equation x² + 3x + 1 to the standard equation ax² + bx + c, we get a = 1, b = 3, and c = 1.
Now, the nature of the roots depends upon the discriminant (D),
D = b² - 4ac = 3² - 4.1.1 = 9 - 4 = 5.
∵ Discriminant (D) > 0, the quadratic equation, y = x² + 3x + 1 has both its root real and distinct.
∴ The quadratic equation, y = x² + 3x + 1 has two real roots.
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