# Square root of x^2-5x+4 is defined for x = ?

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### 2 Answers

The square root of x^2 - 5x + 4 is a real value if x^2 - 5x + 4 >=0

x^2 - 5x + 4 = 0

=> x^2 - 4x - x + 4 =0

=> x(x - 4) - 1(x - 4) = 0

=> (x - 1)(x - 4) = 0

The roots are x = 1 and x = 4

For values of x such that 1< x < 4, x^2 - 5x + 4 < 0

Therefore the square root is defined when

**x <= 1 and x >= 4.**

The square root exists if and only if only the expression x^2-5x+4 is positive or zero.

To determine the range of x values for the expression x^2-5x+4 to be positive, we'll have to calculate the roots of the expression x^2-5x+4.

x1=[-(-5)+sqrt(25-16)]/2

x1=(5+3)/2

x1=4

x2=[-(-5)-sqrt(25-16)]/2

x2=(5-3)/2

x2=1

The radicand is positive if x belongs to the ranges (-infinite,1] or [4,+infinite) and it is negative for (1,4).

**So, the square root is defined if x belongs to the reunion of intervals: (-infinite,1] U [4,+infinite).**