Solve a-1=sqrt{7-a}

$$a - 1 = \sqrt { 7 - a }$$

Solve for a

$$a=3$$

Solution Steps

Square both sides of the equation.

$$\left(a-1\right)^{2}=\left(\sqrt{7-a}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$a^{2}-2a+1=\left(\sqrt{7-a}\right)^{2}$$

Calculate \sqrt{7-a} to the power of 2 and get 7-a.

$$a^{2}-2a+1=7-a$$

Subtract 7 from both sides.

$$a^{2}-2a+1-7=-a$$

Subtract 7 from 1 to get -6.

$$a^{2}-2a-6=-a$$

Add a to both sides.

$$a^{2}-2a-6+a=0$$

Combine -2a and a to get -a.

$$a^{2}-a-6=0$$

To solve the equation, factor a^{2}-a-6 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-3\right)\left(a+2\right)$$

To find equation solutions, solve a-3=0 and a+2=0.

$$a=3 a=-2$$

Substitute 3 for a in the equation a-1=\sqrt{7-a}.

$$3-1=\sqrt{7-3}$$

Simplify. The value a=3 satisfies the equation.

$$2=2$$

Substitute -2 for a in the equation a-1=\sqrt{7-a}.

$$-2-1=\sqrt{7-\left(-2\right)}$$

Simplify. The value a=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

Equation a-1=\sqrt{7-a} has a unique solution.

$$a=3$$