Question

8.6 solve radical equations (homework)
score: 11/16 answered: 11/16
question 12
solve: \\(\sqrt{y + 49}-y=-7\\)
\\(y =\\)

Explanation:

Step1: Isolate the radical term

Add \(y\) to both sides of the equation \(\sqrt{y + 49}-y=-7\) to get \(\sqrt{y + 49}=y - 7\).

Step2: Square both sides

\((\sqrt{y + 49})^2=(y - 7)^2\), which simplifies to \(y + 49=y^{2}-14y + 49\).

Step3: Rearrange to form a quadratic equation

Subtract \(y+49\) from both sides: \(0=y^{2}-14y + 49-(y + 49)\), resulting in \(y^{2}-15y=0\).

Step4: Factor the quadratic equation

Factor out \(y\): \(y(y - 15)=0\).

Step5: Solve for \(y\)

Set each factor equal to zero. If \(y=0\), then \(\sqrt{0 + 49}-0=7
eq - 7\), so \(y = 0\) is an extraneous solution. If \(y-15=0\), then \(y = 15\). Checking \(y = 15\), \(\sqrt{15+49}-15=\sqrt{64}-15=8 - 15=-7\).

Answer:

\(y = 15\)