Find the value of k for which the quadratic equation 2x^2-kx+k=0 has equal real roots
Answers
Real roots =>b^2-4ac=0
(-k)^2-4(2)(k)=0
k^2-8k=0
k^2=8k
k=8k÷k
k=8
Question:
Find the value of k for which the quadratic equation 2x² -kx + k= 0 has equal roots.
Answer:
k = 0 or 8
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 + k = 0 .
Clearly , we have ;
a = 2
b = -k
c = k
We know that ,
The quadratic equation will have real and equal roots if its discriminant is zero .
=> D = 0
=> (-k)² - 4•2•k = 0
=> k² - 8k = 0
=> k•(k-8) = 0
=> k = 0 or 8
Hence,
The required values of k are 0 and 8 .