Solve (4 + sqrt(2))/(2 + sqrt(2)) = a - sqrt(b) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$$

Solve for b

$b=\left(-a+3-\sqrt{2}\right)^{2}$
$-\left(-a+3-\sqrt{2}\right)\geq 0$

Steps for Solving Radical Equations

Rationalize the denominator of $\frac{4+\sqrt{2}}{2+\sqrt{2}}$ by multiplying numerator and denominator by $2-\sqrt{2}$.

$$\frac{\left(4+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=a-\sqrt{b}$$

Consider $\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(4+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2^{2}-\left(\sqrt{2}\right)^{2}}=a-\sqrt{b}$$

Square $2$. Square $\sqrt{2}$.

$$\frac{\left(4+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{4-2}=a-\sqrt{b}$$

Subtract $2$ from $4$ to get $2$.

$$\frac{\left(4+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=a-\sqrt{b}$$

Use the distributive property to multiply $4+\sqrt{2}$ by $2-\sqrt{2}$ and combine like terms.

$$\frac{8-2\sqrt{2}-\left(\sqrt{2}\right)^{2}}{2}=a-\sqrt{b}$$

The square of $\sqrt{2}$ is $2$.

$$\frac{8-2\sqrt{2}-2}{2}=a-\sqrt{b}$$

Subtract $2$ from $8$ to get $6$.

$$\frac{6-2\sqrt{2}}{2}=a-\sqrt{b}$$

Divide each term of $6-2\sqrt{2}$ by $2$ to get $3-\sqrt{2}$.

$$3-\sqrt{2}=a-\sqrt{b}$$

Swap sides so that all variable terms are on the left hand side.

$$a-\sqrt{b}=3-\sqrt{2}$$

Subtract $a$ from both sides.

$$-\sqrt{b}=3-\sqrt{2}-a$$

Divide both sides by $-1$.

$$\frac{-\sqrt{b}}{-1}=\frac{-a+3-\sqrt{2}}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$\sqrt{b}=\frac{-a+3-\sqrt{2}}{-1}$$

Divide $3-\sqrt{2}-a$ by $-1$.

$$\sqrt{b}=a+\sqrt{2}-3$$

Square both sides of the equation.

$$b=\left(a+\sqrt{2}-3\right)^{2}$$

Solve for a

$a=\sqrt{b}+3-\sqrt{2}$
$b\geq 0$