How do you solve $\sqrt{x^{2} + 7} = x + 1$ $sqrt( x^2+7)=x+1$ ?

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How do you solve $\sqrt{x^{2} + 7} = x + 1$ $sqrt( x^2+7)=x+1$ ?

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Answer 1 · 2016-05-08T12:22:59

$x = 3$ $x=3$

Explanation:

First square both sides, noting that squaring may introduce spurious solutions...

$x^{2} + 7 = {(x + 1)}^{2} = x^{2} + 2 x + 1$ $x^2+7 = (x+1)^2 = x^2+2x+1$

Subtract $x^{2}$ $x^2$ from both ends to get:

$7 = 2 x + 1$ $7 = 2x+1$

Subtract $1$ $1$ from both sides to get:

$6 = 2 x$ $6 = 2x$

Divide both sides by $2$ $2$ and transpose to get:

$x = 3$ $x = 3$

Check that this is a solution of the original equation:

$\sqrt{x^{2} + 7} = \sqrt{3^{2} + 7} = \sqrt{9 + 7} = \sqrt{16} = 4 = 3 + 1 = x + 1$ $sqrt(x^2+7) = sqrt(3^2+7) = sqrt(9+7) = sqrt(16) = 4 = 3+1 = x+1$

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How do you solve $\sqrt{x^{2} + 7} = x + 1$ $sqrt( x^2+7)=x+1$ ?

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How do you solve $\sqrt{x^{2} + 7} = x + 1$ $sqrt( x^2+7)=x+1$ ?