Solve for $b$ . $\newline$ $\sqrt{b+1}=\sqrt{b+6}-1$

View step-by-step help

Home

Math Problems

Algebra 1

Multiply radical expressions

Full solution

Q. Solve for

b

\newline

\sqrt{b+1}=\sqrt{b+6}-1

Isolate $b$ in equation: We are given the equation $\sqrt{b+1} = \sqrt{b+6} - 1$ . To solve for $b$ , we need to isolate $b$ on one side of the equation.
Square both sides: First, let's get rid of the square roots by squaring both sides of the equation. This will eliminate the square root on the left side and we will have to square the entire right side of the equation. $\newline$ (\sqrt{b+$1$})^\(2 = (\sqrt{b+ $6$ } - $1$ )^ $2$
Apply binomial square formula: Squaring the left side gives us $b + 1$ . On the right side, we need to apply the binomial square formula $(a - b)^2 = a^2 - 2ab + b^2$ . $\newline$ $b + 1 = (\sqrt{b+6})^2 - 2\cdot\sqrt{b+6}\cdot 1 + 1^2$
Simplify right side: Squaring $\sqrt{b+6}$ gives us $b + 6$ . We also simplify the rest of the right side of the equation. $b + 1 = b + 6 - 2\sqrt{b+6} + 1$
Combine like terms: Now, let's simplify the right side by combining like terms. $b + 1 = b + 7 - 2\sqrt{b+6}$
Subtract $b$ from both sides: Next, we subtract $b$ from both sides to get the terms with $b$ on one side and the constant terms on the other side. $b + 1 - b = b + 7 - b - 2\sqrt{b+6}$
Isolate term with square root: After simplifying, we have: $\newline$ $1 = 7 - 2\sqrt{b+6}$
Divide both sides by $-2$ : Now, we subtract $7$ from both sides to isolate the term with the square root. $\newline$ $1 - 7 = -2\sqrt{b+6}$
Square both sides again: This simplifies to: $\newline$ $-6 = -2\sqrt{b+6}$
Solve for b: Next, we divide both sides by $-2$ to solve for $\sqrt{b+6}$ . $-6 / -2 = -2\cdot\sqrt{b+6} / -2$
Solve for b: Next, we divide both sides by $-2$ to solve for $\sqrt{b+6}$ . $-6 / -2 = -2\cdot\sqrt{b+6} / -2$ This gives us: $3 = \sqrt{b+6}$
Solve for b: Next, we divide both sides by $-2$ to solve for $\sqrt{b+6}$ . $-6 / -2 = -2\cdot\sqrt{b+6} / -2$ This gives us: $3 = \sqrt{b+6}$ Now, we square both sides again to eliminate the square root and solve for b. $3^2 = (\sqrt{b+6})^2$
Solve for b: Next, we divide both sides by $-2$ to solve for $\sqrt{b+6}$ . $-6 / -2 = -2\cdot\sqrt{b+6} / -2$ This gives us: $3 = \sqrt{b+6}$ Now, we square both sides again to eliminate the square root and solve for $b$ . $3^2 = (\sqrt{b+6})^2$ Squaring $3$ gives us $9$ , and squaring $\sqrt{b+6}$ gives us $b + 6$ . $\sqrt{b+6}$ $0$
Solve for b: Next, we divide both sides by $-2$ to solve for $\sqrt{b+6}$ .
$-6 / -2 = -2\cdot\sqrt{b+6} / -2$ This gives us:
$3 = \sqrt{b+6}$ Now, we square both sides again to eliminate the square root and solve for $b$ .
$3^2 = (\sqrt{b+6})^2$ Squaring $3$ gives us $9$ , and squaring $\sqrt{b+6}$ gives us $b + 6$ .
$\sqrt{b+6}$ $0$ Finally, we subtract $\sqrt{b+6}$ $1$ from both sides to solve for $b$ .
$\sqrt{b+6}$ $3$
Solve for b: Next, we divide both sides by $-2$ to solve for $\sqrt{b+6}$ .
$-6 / -2 = -2\cdot\sqrt{b+6} / -2$ This gives us:
$3 = \sqrt{b+6}$ Now, we square both sides again to eliminate the square root and solve for $b$ .
$3^2 = (\sqrt{b+6})^2$ Squaring $3$ gives us $9$ , and squaring $\sqrt{b+6}$ gives us $b + 6$ .
$\sqrt{b+6}$ $0$ Finally, we subtract $\sqrt{b+6}$ $1$ from both sides to solve for $b$ .
$\sqrt{b+6}$ $3$ This gives us the final answer:
$\sqrt{b+6}$ $4$