find the value of K for which the given equation has equal roots (k-12) x^2+2(k-12)x+2=0 About the author Eden
Answer: The values of k for which the given equation has equal roots are 14 and 12. Step-by-step explanation: Given polynomial is; (k-12) x^2+2(k-12)x+2=0; here, a = (k-12); b = -2(k-12); c = 2 For equal roots, we know a discriminant and that is; b² – 4ac = 0; Thus , (-2(k-12))² – 4(k-12)(2) = 0; 4(k-12)²- 4(k-12)(2) = 0; First value of k will be; 4(k-12){(k-12) – 2} = 0; (k-12) – 2 = 0 / 4(k-12); k – 12 – 2 = 0; k – 14= 0; k = 14. Second value of k will be; 4(k-12){(k-12) – 2} = 0; 4(k-12){(k-12) – 2} = 0 / {(k-12) – 2}; 4(k-12) = 0; k – 12= 0; k = 12. That’s all. Reply
Answer:
The values of k for which the given equation has equal roots are 14 and 12.
Step-by-step explanation:
Given polynomial is;
(k-12) x^2+2(k-12)x+2=0;
here, a = (k-12);
b = -2(k-12);
c = 2
For equal roots, we know a discriminant and that is;
b² – 4ac = 0;
Thus , (-2(k-12))² – 4(k-12)(2) = 0;
4(k-12)²- 4(k-12)(2) = 0;
First value of k will be;
4(k-12){(k-12) – 2} = 0;
(k-12) – 2 = 0 / 4(k-12);
k – 12 – 2 = 0;
k – 14= 0;
k = 14.
Second value of k will be;
4(k-12){(k-12) – 2} = 0;
4(k-12){(k-12) – 2} = 0 / {(k-12) – 2};
4(k-12) = 0;
k – 12= 0;
k = 12.
That’s all.