How do you solve $2 x + \sqrt{x + 1} = 8$ $2x + sqrt(x+1) = 8$ ?

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How do you solve $2 x + \sqrt{x + 1} = 8$ $2x + sqrt(x+1) = 8$ ?

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Answer 1 · 2016-03-02T17:15:44

$x = 3$ $x=3$ or $x = 5.25$ $x=5.25$

Explanation:

Subtract $2 x$ $2x$ from bopth sides of the equation, yielding $\sqrt{x + 1} = 8 - 2 x$ $sqrt(x+1) = 8 -2x$ . Square both sides of the equation, yielding $x + 1 = 64 - 32 x + 4 x^{2}$ $x+1=64-32x+4x^2$ . Subtract x+1 from both sides of the equation, yielding $0 = 63 - 33 x + 4 x^{2}$ $0=63-33x+4x^2$ .

Using the quadratic equation, one gets
$x = \frac{33 (\frac{+}{-}) \sqrt{1089 - 4 \cdot 4 \cdot 63}}{8} = \frac{33 (\frac{+}{-}) \sqrt{81}}{8} = \frac{33 (\frac{+}{-}) 9}{8}$ $x = (33 (+/-) sqrt(1089 - 4*4*63))/8 = (33 (+/-) sqrt(81))/8 = (33 (+/-) 9)/8$ .
So, $x = \frac{42}{8}$ $x = 42/8$ or $x = \frac{24}{8}$ $x=24/8$ .

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How do you solve $2 x + \sqrt{x + 1} = 8$ $2x + sqrt(x+1) = 8$ ?

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