Question

fva 25-28: algebra ii q3
solving equations containing two radicals
solving equations containing two radicals
what are the solutions of $sqrt{2x+3}-sqrt{x+1}=1$? choose two correct answers.
-3
-1
3
1

Explanation:

Step1: Isolate one radical

$\sqrt{2x+3} = 1 + \sqrt{x+1}$

Step2: Square both sides

$(\sqrt{2x+3})^2 = (1 + \sqrt{x+1})^2$
$2x+3 = 1 + 2\sqrt{x+1} + x+1$

Step3: Simplify the equation

$2x+3 = x+2 + 2\sqrt{x+1}$
$x+1 = 2\sqrt{x+1}$

Step4: Square both sides again

$(x+1)^2 = (2\sqrt{x+1})^2$
$x^2+2x+1 = 4(x+1)$

Step5: Rearrange to quadratic form

$x^2+2x+1 -4x -4 = 0$
$x^2-2x-3 = 0$

Step6: Factor the quadratic

$(x-3)(x+1) = 0$
$x=3$ or $x=-1$

Step7: Verify solutions

For $x=3$: $\sqrt{2*3+3}-\sqrt{3+1}=\sqrt{9}-\sqrt{4}=3-2=1$, valid.
For $x=-1$: $\sqrt{2*(-1)+3}-\sqrt{-1+1}=\sqrt{1}-0=1$, valid.
For $x=-3$: $\sqrt{2*(-3)+3}$ is undefined (negative radicand).
For $x=1$: $\sqrt{2*1+3}-\sqrt{1+1}=\sqrt{5}-\sqrt{2}\approx2.236-1.414
eq1$, invalid.

Correction on verification note: The valid solutions are $x=3$ and $x=-1$

Step1: Isolate one radical

$\sqrt{2x+3} = 1 + \sqrt{x+1}$

Step2: Square both sides

$2x+3 = 1 + 2\sqrt{x+1} + x+1$

Step3: Simplify to isolate remaining radical

$x+1 = 2\sqrt{x+1}$

Step4: Square both sides again

$x^2+2x+1 = 4(x+1)$

Step5: Form quadratic equation

$x^2-2x-3=0$

Step6: Solve quadratic

$(x-3)(x+1)=0 \implies x=3, x=-1$

Step7: Validate solutions

• $x=3$: $\sqrt{9}-\sqrt{4}=3-2=1$ ✔️
• $x=-1$: $\sqrt{1}-\sqrt{0}=1-0=1$ ✔️
• $x=-3$: $\sqrt{-3}$ undefined ❌
• $x=1$: $\sqrt{5}-\sqrt{2}

eq1$ ❌

Answer:

3, 1