Solve (sqrt(a) + sqrt(b))(sqrt(b) - sqrt(a)) = (sqrt(b) + sqrt(c))(sqrt(c) - 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

$$(\sqrt{a}+\sqrt{b})(\sqrt{b}-\sqrt{a})=(\sqrt{b}+\sqrt{c})(\sqrt{c}-\sqrt{b})$$

Solve for a

$\left\{\begin{matrix}a=0\text{, }&c=2b\text{ and }b\geq 0\\a=2b-c\text{, }&b\geq \frac{c}{2}\text{ and }c\geq 0\end{matrix}\right.$

Steps for Solving Linear Equation

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

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

Calculate $\sqrt{b}$ to the power of $2$ and get $b$.

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

Calculate $\sqrt{a}$ to the power of $2$ and get $a$.

$$b-a=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{c}-\sqrt{b}\right)$$

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

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

Calculate $\sqrt{c}$ to the power of $2$ and get $c$.

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

Calculate $\sqrt{b}$ to the power of $2$ and get $b$.

$$b-a=c-b$$

Subtract $b$ from both sides.

$$-a=c-b-b$$

Combine $-b$ and $-b$ to get $-2b$.

$$-a=c-2b$$

Divide both sides by $-1$.

$$\frac{-a}{-1}=\frac{c-2b}{-1}$$

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

$$a=\frac{c-2b}{-1}$$

Divide $-2b+c$ by $-1$.

$$a=2b-c$$

Solve for b

$\left\{\begin{matrix}b=0\text{, }&a=0\text{ and }c=0\\b=\frac{a+c}{2}\text{, }&a\geq 0\text{ and }c\geq 0\end{matrix}\right.$

Steps for Solving Linear Equation

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

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

Calculate $\sqrt{b}$ to the power of $2$ and get $b$.

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

Calculate $\sqrt{a}$ to the power of $2$ and get $a$.

$$b-a=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{c}-\sqrt{b}\right)$$

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

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

Calculate $\sqrt{c}$ to the power of $2$ and get $c$.

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

Calculate $\sqrt{b}$ to the power of $2$ and get $b$.

$$b-a=c-b$$

Add $b$ to both sides.

$$b-a+b=c$$

Combine $b$ and $b$ to get $2b$.

$$2b-a=c$$

Add $a$ to both sides.

$$2b=c+a$$

The equation is in standard form.

$$2b=a+c$$

Divide both sides by $2$.

$$\frac{2b}{2}=\frac{a+c}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$b=\frac{a+c}{2}$$