Find the value of k for which the quadratic equation 2x^2+kx+3=0 has equal roots
Answers
b2−4ac=0
k2−4×2×3=0
k2−24=0
k2=24k
=−+√24k
K=+−2√6
Question:
Find the value of k for which the quadratic equation 2x²+ kx + 3=0 has equal roots.
Answer:
k = ± 2√6
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 ;
2x² + kx + 3 = 0 .
Clearly , we have ;
a = 2
b = k
c = 3
We know that ,
The quadratic equation will have real and equal roots if its discriminant is zero .
=> D = 0
=> b² - 4ac = 0
=> k² - 4•2•3 = 0
=> k² - 24 = 0
=> k² = 24
=> k = √24
=> k = ± 2√6
Hence,
The required values of k are ± 2√6 .