find the value of k for which the quadratic equation x^+2√2kx+18=0
Answers
Answer:
8
Step-by-step explanation:
x2 is the fact of polynomial
Question:
Find the value of k for which the quadratic equation x² + 2√2kx + 18 = 0 has equal roots.
Answer:
k = ± 3
Note:
• An equation of degree 2 is know as quadratic equation .
• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.
• The maximum number of roots of an equation will be equal to its degree.
• A quadratic equation has atmost two roots.
• The general form of a quadratic equation is given as , ax² + bx + c = 0 .
• A quadratic equation has atmost two roots.
• The general form of a quadratic equation is given as , ax² + bx + c = 0 .
• The discriminant of the quadratic equation is given as , D = b² - 4ac .
• If D = 0 , then the quadratic equation would have real and equal roots .
• If D > 0 , then the quadratic equation would have real and distinct roots .
• If D < 0 , then the quadratic equation would have imaginary roots .
Solution:
The given quadratic equation is ;
x² + 2√2kx + 18 = 0 .
Clearly , we have ;
a = 1
b = 2√2k
c = 18
We know that ,
The quadratic equation will have real and equal roots if its discriminant is zero .
=> D = 0
=> (2√2k)² - 4•1•18 = 0
=> 8k² - 4•18 = 0
=> 8•(k² - 9) = 0
=> k² - 9 = 0
=> k² = 9
=> k = √9
=> k = ± 3
Hence,
The required values of k are ± 3 .