How do you solve $\sqrt{x - 1} = x - 3$ $sqrt(x -1) = x - 3$ ?

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Answer 1 · 2018-04-25T23:15:24

x=2,5

To remove the square root, square both sides

$x - 1 = {(x - 3)}^{2}$ $x-1=(x-3)^2$

Square the right side

${(x - 3)}^{2} = (x - 3) (x - 3) = x^{2} - 3 x - 3 x + 9$ $(x-3)^2=(x-3)(x-3)=x^2-3x-3x+9$

$x - 1 = x^{2} - 6 x + 9$ $x-1=x^2-6x+9$

Combine terms and set equation to 0

$0 = x^{2} - 7 x + 10$ $0=x^2-7x+10$

Factor

$0 = (x - 2) (x - 5)$ $0=(x-2)(x-5)$

How do you solve √x−1=x−3sqrt(x -1) = x - 3?