.Find the value of k for which the roots of the quadratic equation (k – 5)x2 + 2(k – 5) x + 2 = 0 are equal.
Answers
The quadratic equation (k-5)x2+2(k-5)x+2=0
If roots are equal, then
B^2-4AC=0
4(k-5)^2-8(k-5)=0
4(k^2-10k+25)-8k+40=0
4k^2-40k+100-8k+40=0
4k^2-48k+140=0
K^2-12k+35=0
K^2-7k-5k+35=0
K(k-7)-5(k-7)=0
(k-7)(k-5)=0
So,k=5,7.
Question:
Find the value of k for which the quadratic equation (k-5)x² + 2(k-5)x + 2 = 0 has equal roots.
Answer:
k = 5 , 7
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 .
• 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 ;
(k-5)x² + 2(k-5)x + 2 = 0
Clearly , we have ;
a = k-5
b = 2(k-5)
c = 2
We know that ,
The quadratic equation will have equal roots if its discriminant is equal to zero .
=> D = 0
=> [2(k-5)]² - 4•(x-5)•2 = 0
=> 4(k-5)² - 4•2(k-5) = 0
=> 4(k-5)•(k-5-2) = 0
=> (k-5)(k-7) = 0
=> k = 5 , 7