Solve k^2=0.49

$$k ^ { 2 } = 0.49$$

Solve for k

$$k=0.7$$

$$k=-0.7$$

Steps Using Difference of Squares

Subtract 0.49 from both sides.

$$k^{2}-0.49=0$$

Consider k^{2}-0.49. Rewrite k^{2}-0.49 as k^{2}-\left(\frac{7}{10}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(k-\frac{7}{10}\right)\left(k+\frac{7}{10}\right)=0$$

To find equation solutions, solve k-\frac{7}{10}=0 and k+\frac{7}{10}=0.

$$k=\frac{7}{10} k=-\frac{7}{10}$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$k=\frac{7}{10} k=-\frac{7}{10}$$

Steps Using the Quadratic Formula

Subtract 0.49 from both sides.

$$k^{2}-0.49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -0.49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\left(-0.49\right)}}{2}$$

Square 0.

$$k=\frac{0±\sqrt{-4\left(-0.49\right)}}{2}$$

Multiply -4 times -0.49.

$$k=\frac{0±\sqrt{1.96}}{2}$$

Take the square root of 1.96.

$$k=\frac{0±\frac{7}{5}}{2}$$

Now solve the equation k=\frac{0±\frac{7}{5}}{2} when ± is plus.

$$k=\frac{7}{10}$$

Now solve the equation k=\frac{0±\frac{7}{5}}{2} when ± is minus.

$$k=-\frac{7}{10}$$

The equation is now solved.

$$k=\frac{7}{10} k=-\frac{7}{10}$$