Solve 4b^2-121=0

$$4 b ^ { 2 } - 121 = 0$$

Solve for b

$$b = \frac{11}{2} = 5\frac{1}{2} = 5.5$$

$$b = -\frac{11}{2} = -5\frac{1}{2} = -5.5$$

Steps Using Difference of Squares

Consider 4b^{2}-121. Rewrite 4b^{2}-121 as \left(2b\right)^{2}-11^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2b-11\right)\left(2b+11\right)=0$$

To find equation solutions, solve 2b-11=0 and 2b+11=0.

$$b=\frac{11}{2} b=-\frac{11}{2}$$

Steps by Finding Square Root

Add 121 to both sides. Anything plus zero gives itself.

$$4b^{2}=121$$

Divide both sides by 4.

$$b^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$b=\frac{11}{2} b=-\frac{11}{2}$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$4b^{2}-121=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -121 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times 4\left(-121\right)}}{2\times 4}$$

Square 0.

$$b=\frac{0±\sqrt{-4\times 4\left(-121\right)}}{2\times 4}$$

Multiply -4 times 4.

$$b=\frac{0±\sqrt{-16\left(-121\right)}}{2\times 4}$$

Multiply -16 times -121.

$$b=\frac{0±\sqrt{1936}}{2\times 4}$$

Take the square root of 1936.

$$b=\frac{0±44}{2\times 4}$$

Multiply 2 times 4.

$$b=\frac{0±44}{8}$$

Now solve the equation b=\frac{0±44}{8} when ± is plus. Reduce the fraction \frac{44}{8} to lowest terms by extracting and canceling out 4.

$$b=\frac{11}{2}$$

Now solve the equation b=\frac{0±44}{8} when ± is minus. Reduce the fraction \frac{-44}{8} to lowest terms by extracting and canceling out 4.

$$b=-\frac{11}{2}$$

The equation is now solved.

$$b=\frac{11}{2} b=-\frac{11}{2}$$