Solve the following equation: bb−5−2b+5=50b2−25 and show your work.(4 points)

Answer 1

To solve the equation $bb-5-2b+5=50b^2-25$, we first rewrite $bb$ as $b^2$. Thus, the equation becomes:

$$b^2-5-2b+5=50b^2-25$$

Now, simplify the left side:

$$b^2-2b+0=50b^2-25$$

This simplifies to:

$$b^2-2b=50b^2-25$$

Next, we move all terms to one side of the equation. Subtract $50b^2-25$ from both sides:

$$b^2-2b-50b^2+25=0$$

Combining like terms gives:

$$-b^2-2b+25=0$$

To make it standard, multiply the entire equation by -1:

$$b^2+2b-25=0$$

Now we can solve this quadratic equation using the quadratic formula:

$$b=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$

Where $A=1$, $B=2$, and $C=-25$. Calculating the discriminant:

$$B^2-4AC=2^2-4\cdot 1\cdot (-25)=4+100=104$$

Now substitute back into the formula:

$$b=\frac{-2\pm \sqrt{104}}{2\cdot 1}$$

$\sqrt{104}$ can be simplified to $2\sqrt{26}$, thus we have:

$$b=\frac{-2\pm 2\sqrt{26}}{2}$$ $$b=-1\pm \sqrt{26}$$

Hence, the two possible solutions for $b$ are:

$$b=-1+\sqrt{26}\text{and}b=-1-\sqrt{26}$$

Therefore, the final solutions are:

$$\boxed{{\displaystyle -1+\sqrt{26}}}\text{and}\boxed{{\displaystyle -1-\sqrt{26}}}$$