Find "a" in this equation.
Answers
Answer:
the value of a in the equation is 2
Hello !!
For you answer this question, you only need make use of this method below. See the resolution.
5 = √[(5 - a)^2 + (3 - (-1))^2]
5 = √[(5 - a)(5 - a) + (3 - (-1))(3 - (-1))]
5 = √[25 - 10a + a^2 + 16]
5 = √[-10a + a^2 + 16 + 25]
5 = √[- 10a + a^2 + 41]
√[- 10a + a^2 + 41] = 5
[- 10a + a^2 + 41] = 5^2
-10a + a^2 + 41 = 25
-10a + a^2 + 41 - 25 = 0
-10a + a^2 + 16 = 0
a^2 - 10a + 16 = 0
We have a quadratic equation, now we go find the two solution for a.
Equation :
a^2 - 10a + 16 = 0
Coeficients.
A = 1 ; B = -10 ; C = 16
Discriminant.
D = b^2 - 4ac
D = (-10)^2 - 4(1)(16)
D = 100 - 64
D = 36
Find the roots.
a = (-b ± √(D))/(2a)
a = (-(-10) ± √(36))/(2 × 1)
a = (10 ± 6)/2
a' = (10 + 6)/2
a' = 16/2
a' = 8
a'' = (10 - 6)/2
a'' = 4/2
a'' = 2
We have two solution for (a). For you give a comprovation you can check it.
Give comprovation for the first solution (a = 8).
5 = √[(5 - a)^2 + (3 - (-1))^2]
5 = √[(5 - 8)^2 + (3 - (-1))^2]
5 = √[(-3)^2 + (3 + 1))^2]
5 = √[(-3)^2 + (4))^2]
5 = √[9 + 16]
5 = √25
5 = 5 (TRUE).
Give comprovation for the second solution (a = 2).
5 = √[(5 - a)^2 + (3 - (-1))^2]
5 = √[(5 - 2)^2 + (3 - (-1))^2]
5 = √[(3)^2 + (3 + 1))^2]
5 = √[(3)^2 + (4)^2]
5 = √[9 + 16]
5 = √25
5 = 5 (TRUE).
Final result : has two solution for (a), are 8 and 2.
I hope I have collaborated !