find the value of k for the following quadratic equation so that it has two real and equal roots 5x²-2kx+20 =0
Answers
Answer:
5x2−2kx+20=0
It is given that the roots of the quadratic equation are real and equal, Then discriminant D=0.
Comparing the given equation with ax2+bx+c=0
we have a=5,b=−2k and c=20
Now, D=0
∴b2−4ac=0
∴(−2k)2−4(5)(20)=0
∴4k2−400=0
∴k2=100
∴k=±10
Thus, k=10 or −10
Step-by-step explanation:
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Given : 5x²-2kx+20 =0
two real and equal roots
To Find : value of k
Solution:
Let say Equal roots are
α , α
Product of roots = α. α = 20/5
=> α² = 4
=> α = ± 2
Sum of roots = α + α = 2α
= 2 ( ± 2)
= ± 4
Sum of roots = - (-2k)/5
= 2k/5
2k/5 = ± 4
=> k = ± 10
Value of k = ± 10
Another method :
Roots are equal if D = 0
=> (-2k)² - 4(5)(20) = 0
=> 4k² - 400 = 0
=> k² - 100 = 0
=> (k + 10)(k - 10) = 0
=> k = 10 , - 10
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