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if the roots of x²-7x+k=0 are m and m-1, find the value of the constant k​

Sagot :

zanescott49

Answer: k=12

Step-by-step explanation:

x2 - 7x + k = 0

using quadratic formula method,

x = [-b ± √(b2 - 4ac)]/2a

a = 1

b = -7

c = k

x = [-(-7) ± √((-7)2 - 4(1)(k))]/2(1)

x = [7 ± √(49 - 4k)]/2

x = [7 + √(49 - 4k)] / 2 and [7 - √(49 - 4k)] / 2

since the roots are m and m-1,

m = [7 + √(49 - 4k)] / 2 ------------ eqn(1), and

m - 1 = [7 - √(49 - 4k)] / 2 ----------- eqn(2)

add 1 to both sides of eqn(2)

m = [7 - √(49 - 4k)]/2 + 1 --------------- eqn(3)

eqn(1) = eqn(3)

[7 + √(49 - 4k)]/2 = [7 - √(49 - 4k)]/2 + 1

7/2 + (√(49 - 4k))/2 = 7/2 - (√(49 - 4k))/2 + 1

collect like terms,

(√(49 - 4k))/2 + (√(49 - 4k))/2 = 7/2 - 7/2 + 1

(√(49 - 4k))/2 + (√(49 - 4k))/2 = 0 + 1

(√(49 - 4k))/2 + (√(49 - 4k))/2 = 1

multiply through by 2

√(49 - 4k) + √(49 - 4k) = 2

2√(49 - 4k) = 2

divide both side by 2

√(49 - 4k) = 1

square both sides

49 - 4k = 1

collect like terms,

4k = 49 - 1

4k = 48

divide both sides by 4

k = 48/4

k = 12

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