Find the positive value of k for which the equation 2x2 + kx + 2 = 0 has real root
Answers
Solution:-
:-Given equation
=> 2x² + kx + 2 = 0
:-Now compare with
=> ax² + bx + c = 0
We get
=> a = 2 , b = k and c = 2
Formula to find discriminant
=> D = b² - 4ac
Put the value
=> D = k² - 4 × 2 × 2
=> D = k² - 16
Its given:- for real roots D = 0
=> k² - 16 = 0
=> k² = 16
=> k = √16
=> k = ± 4
So value of k is + 4 and - 4
More about quadratic equation
Quadratic equation, in mathematics, an algebraic equation of the second degree (having one or more variables raised to the second power). Old Babylonian cuneiform texts, dating from the time of Hammurabi, show a knowledge of how to solve quadratic equations, but it appears that ancient Egyptian mathematicians did not know how to solve them. Since the time of Galileo, they have been important in the physics of accelerated motion, such as free fall in a vacuum. The general quadratic equation in one variable is ax2 + bx + c = 0, in which a, b, and c are arbitrary constants (or parameters) and a is not equal to 0. Such an equation has two roots (not necessarily distinct), as given by the quadratic formula
Given that
2x² + kx + 2 =0 and also given that It has real roots Find value of k
We can find the value of x by using Discriminant of QE
So a=2, b=k, c =2
b² - 4ac =0
k²-4(2)(2) =0
k²-16 =0
k = +4 ,-4
Why we taken Discriminant =0?
If the Roots are real Then Discriminant should equal to 0
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DISCRIMINANT IS REPRESENTED BY D
If D=0 roots are real and equal
D >0 roots are Real and distinct
D<0 Roots are complex and conjugate