Question

solve for u, where u is a real number.\\(\sqrt{-8u + 49} = u - 5\\)\
if there is more than one solution, separate them with commas.\
if there is no solution, click on
o solution\.\
u = \boxed{}

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{-8u + 49}=u - 5\), we first square both sides. Squaring the left side gives \((\sqrt{-8u + 49})^2=-8u + 49\), and squaring the right side gives \((u - 5)^2\). Using the formula \((a - b)^2=a^2-2ab + b^2\), we have \((u - 5)^2=u^2-10u + 25\). So the equation becomes:

$$ -8u + 49=u^2-10u + 25 $$

Step2: Rearrange into standard quadratic form

Move all terms to one side to form a quadratic equation. Subtract \(-8u + 49\) from both sides:

$$ u^2-10u + 25+8u - 49 = 0 $$

Simplify the like terms:

$$ u^2-2u - 24 = 0 $$

Step3: Factor the quadratic equation

We need to factor \(u^2-2u - 24\). We look for two numbers that multiply to \(-24\) and add up to \(-2\). The numbers are \(-6\) and \(4\) because \(-6\times4=-24\) and \(-6 + 4=-2\). So the factored form is:

$$ (u - 6)(u + 4)=0 $$

Step4: Solve for u using the zero - product property

If \((u - 6)(u + 4)=0\), then either \(u - 6 = 0\) or \(u + 4 = 0\). Solving \(u - 6 = 0\) gives \(u = 6\). Solving \(u + 4 = 0\) gives \(u=-4\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• For \(u = 6\):

Left side: \(\sqrt{-8\times6 + 49}=\sqrt{-48 + 49}=\sqrt{1}=1\)
Right side: \(6 - 5 = 1\)
Since the left side equals the right side, \(u = 6\) is a valid solution.

• For \(u=-4\):

Left side: \(\sqrt{-8\times(-4)+49}=\sqrt{32 + 49}=\sqrt{81}=9\)
Right side: \(-4 - 5=-9\)
Since \(9
eq - 9\), \(u=-4\) is an extraneous solution.

Answer:

\(6\)