Question

8.6 solve radical equations (homework)
score: 10/16 answered: 10/16
question 11
solve: $sqrt{b + 100}-b=-10$
b =
question help: video written example
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Explanation:

Step1: Isolate the radical term

Add \(b\) to both sides of the equation \(\sqrt{b + 100}-b=-10\) to get \(\sqrt{b + 100}=b - 10\).

Step2: Square both sides

Squaring both sides gives \((\sqrt{b + 100})^2=(b - 10)^2\), which simplifies to \(b + 100=b^{2}-20b + 100\).

Step3: Rearrange to form a quadratic equation

Subtract \(b+100\) from both sides: \(0=b^{2}-20b + 100-(b + 100)\), resulting in \(b^{2}-21b=0\).

Step4: Factor the quadratic equation

Factor out \(b\): \(b(b - 21)=0\).

Step5: Solve for \(b\)

Setting each factor equal to zero gives \(b = 0\) or \(b=21\).

Step6: Check for extraneous solutions

For \(b = 0\), \(\sqrt{0 + 100}-0=10
eq-10\), so \(b = 0\) is an extraneous solution.
For \(b = 21\), \(\sqrt{21+100}-21=\sqrt{121}-21=11 - 21=-10\).

Answer:

\(21\)