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Medal Solve sqrtx-1=x-3 check for extraneous solutions A. No solution B. x=5 C. x=2 D. x=2, 5
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is it \[\sqrt{x}-1=x-3\] or \[\sqrt{x-1}=x-3\]??
the second one
Square both sides. then you will get a quadratic equation. Solution to Q.E of the form is \[a^{2}+bx +c=0 \] \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
joseph91 is right
x-1=(x-3)^2 x-1=x^2-6x+9 \[x ^{2}-7x+10=0\]
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solve this equation there are two roots of this equation (2,5) (x-2)(x-5)=0
quadratic equation should be \[ax ^{2}+bx+c=0\] joseph by mistaken wrote \[a ^{2}\] instead of\[ax ^{2}\]
sorry for mistake
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