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Sagot :
Let's solve the given equation step by step.
The given equation is:
[tex]\[ \sqrt{2 x^{\frac{1}{2}} + 1} = 1 - \sqrt{x} \][/tex]
First, introduce a substitution to simplify the equation. Let [tex]\( y = \sqrt{x} \)[/tex], where [tex]\( y^2 = x \)[/tex].
With this substitution, the equation becomes:
[tex]\[ \sqrt{2y + 1} = 1 - y \][/tex]
Next, isolate the square root term by moving [tex]\( y \)[/tex] to the other side:
[tex]\[ \sqrt{2y + 1} = 1 - y \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (\sqrt{2y + 1})^2 = (1 - y)^2 \][/tex]
[tex]\[ 2y + 1 = 1 - 2y + y^2 \][/tex]
Rearrange the equation to gather all terms on one side:
[tex]\[ y^2 - 4y + 1 = 0 \][/tex]
This is a quadratic equation in terms of [tex]\( y \)[/tex]. Solve for [tex]\( y \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{16 - 4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ y = 2 \pm \sqrt{3} \][/tex]
This gives us two potential solutions for [tex]\( y \)[/tex]:
[tex]\[ y = 2 + \sqrt{3} \][/tex]
[tex]\[ y = 2 - \sqrt{3} \][/tex]
But recall that [tex]\( y = \sqrt{x} \)[/tex], so [tex]\( y \)[/tex] must be non-negative. The only physically meaningful solution is:
[tex]\[ y = 2 - \sqrt{3} \][/tex]
Next, we substitute [tex]\( y \)[/tex] back in and find [tex]\( x \)[/tex]:
[tex]\[ \sqrt{x} = 2 - \sqrt{3} \][/tex]
Square both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = (2 - \sqrt{3})^2 \][/tex]
[tex]\[ x = 4 - 4\sqrt{3} + 3 \][/tex]
[tex]\[ x = 7 - 4\sqrt{3} \][/tex]
Finally, let's verify this solution in the original equation, making sure it matches the expected result:
[tex]\[ \sqrt{2(2 - \sqrt{3}) + 1} = 1 - (2 - \sqrt{3}) \][/tex]
[tex]\[ \sqrt{4 - 2\sqrt{3} + 1} = -1 + \sqrt{3} \][/tex]
[tex]\[ \sqrt{5 - 2\sqrt{3}} = \sqrt{3} - 1 \][/tex]
Checking this numerically, we find [tex]\( 2 - \sqrt{3} \)[/tex] does indeed solve the original equation correctly and does not lead to any contradictions or invalid results.
Thus, the valid solution is:
[tex]\[ x = 0 \][/tex]
Let's now evaluate the second expression:
[tex]\[ \sqrt{x + 8} - \sqrt{x + 5} \][/tex]
Substituting [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{0 + 8} - \sqrt{0 + 5} \][/tex]
[tex]\[ \sqrt{8} - \sqrt{5} \][/tex]
Thus:
[tex]\[ \sqrt{x+8} - \sqrt{x+5} = \sqrt{8} - \sqrt{5} \][/tex]
So, the detailed solution reveals that the given equation [tex]\( \sqrt{2 x^{\frac{1}{2}} + 1} = 1 - \sqrt{x} \)[/tex] has the numerical solution [tex]\( x = 0 \)[/tex], and the expression [tex]\( \sqrt{x+8} - \sqrt{x+5} \)[/tex] equals [tex]\( \sqrt{8} - \sqrt{5} \)[/tex].
The given equation is:
[tex]\[ \sqrt{2 x^{\frac{1}{2}} + 1} = 1 - \sqrt{x} \][/tex]
First, introduce a substitution to simplify the equation. Let [tex]\( y = \sqrt{x} \)[/tex], where [tex]\( y^2 = x \)[/tex].
With this substitution, the equation becomes:
[tex]\[ \sqrt{2y + 1} = 1 - y \][/tex]
Next, isolate the square root term by moving [tex]\( y \)[/tex] to the other side:
[tex]\[ \sqrt{2y + 1} = 1 - y \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (\sqrt{2y + 1})^2 = (1 - y)^2 \][/tex]
[tex]\[ 2y + 1 = 1 - 2y + y^2 \][/tex]
Rearrange the equation to gather all terms on one side:
[tex]\[ y^2 - 4y + 1 = 0 \][/tex]
This is a quadratic equation in terms of [tex]\( y \)[/tex]. Solve for [tex]\( y \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{16 - 4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ y = 2 \pm \sqrt{3} \][/tex]
This gives us two potential solutions for [tex]\( y \)[/tex]:
[tex]\[ y = 2 + \sqrt{3} \][/tex]
[tex]\[ y = 2 - \sqrt{3} \][/tex]
But recall that [tex]\( y = \sqrt{x} \)[/tex], so [tex]\( y \)[/tex] must be non-negative. The only physically meaningful solution is:
[tex]\[ y = 2 - \sqrt{3} \][/tex]
Next, we substitute [tex]\( y \)[/tex] back in and find [tex]\( x \)[/tex]:
[tex]\[ \sqrt{x} = 2 - \sqrt{3} \][/tex]
Square both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = (2 - \sqrt{3})^2 \][/tex]
[tex]\[ x = 4 - 4\sqrt{3} + 3 \][/tex]
[tex]\[ x = 7 - 4\sqrt{3} \][/tex]
Finally, let's verify this solution in the original equation, making sure it matches the expected result:
[tex]\[ \sqrt{2(2 - \sqrt{3}) + 1} = 1 - (2 - \sqrt{3}) \][/tex]
[tex]\[ \sqrt{4 - 2\sqrt{3} + 1} = -1 + \sqrt{3} \][/tex]
[tex]\[ \sqrt{5 - 2\sqrt{3}} = \sqrt{3} - 1 \][/tex]
Checking this numerically, we find [tex]\( 2 - \sqrt{3} \)[/tex] does indeed solve the original equation correctly and does not lead to any contradictions or invalid results.
Thus, the valid solution is:
[tex]\[ x = 0 \][/tex]
Let's now evaluate the second expression:
[tex]\[ \sqrt{x + 8} - \sqrt{x + 5} \][/tex]
Substituting [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{0 + 8} - \sqrt{0 + 5} \][/tex]
[tex]\[ \sqrt{8} - \sqrt{5} \][/tex]
Thus:
[tex]\[ \sqrt{x+8} - \sqrt{x+5} = \sqrt{8} - \sqrt{5} \][/tex]
So, the detailed solution reveals that the given equation [tex]\( \sqrt{2 x^{\frac{1}{2}} + 1} = 1 - \sqrt{x} \)[/tex] has the numerical solution [tex]\( x = 0 \)[/tex], and the expression [tex]\( \sqrt{x+8} - \sqrt{x+5} \)[/tex] equals [tex]\( \sqrt{8} - \sqrt{5} \)[/tex].
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