Solve 1/4v_2^2-1/2v_2^2=49/2-49

$$\frac { 1 } { 4 } v _ { 2 } ^ { 2 } - \frac { 1 } { 2 } v _ { 2 } ^ { 2 } = \frac { 49 } { 2 } - 49$$

Solve for v_2

$$v_{2}=7\sqrt{2}\approx 9.899494937$$

$$v_{2}=-7\sqrt{2}\approx -9.899494937$$

Steps by Finding Square Root

Combine \frac{1}{4}v_{2}^{2} and -\frac{1}{2}v_{2}^{2} to get -\frac{1}{4}v_{2}^{2}.

$$-\frac{1}{4}v_{2}^{2}=\frac{49}{2}-49$$

Subtract 49 from \frac{49}{2} to get -\frac{49}{2}.

$$-\frac{1}{4}v_{2}^{2}=-\frac{49}{2}$$

Multiply both sides by -4, the reciprocal of -\frac{1}{4}.

$$v_{2}^{2}=-\frac{49}{2}\left(-4\right)$$

Multiply -\frac{49}{2} and -4 to get 98.

$$v_{2}^{2}=98$$

Take the square root of both sides of the equation.

$$v_{2}=7\sqrt{2} v_{2}=-7\sqrt{2}$$

Steps Using the Quadratic Formula

Combine \frac{1}{4}v_{2}^{2} and -\frac{1}{2}v_{2}^{2} to get -\frac{1}{4}v_{2}^{2}.

$$-\frac{1}{4}v_{2}^{2}=\frac{49}{2}-49$$

Subtract 49 from \frac{49}{2} to get -\frac{49}{2}.

$$-\frac{1}{4}v_{2}^{2}=-\frac{49}{2}$$

Add \frac{49}{2} to both sides.

$$-\frac{1}{4}v_{2}^{2}+\frac{49}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, 0 for b, and \frac{49}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v_{2}=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{4}\right)\times \frac{49}{2}}}{2\left(-\frac{1}{4}\right)}$$

Square 0.

$$v_{2}=\frac{0±\sqrt{-4\left(-\frac{1}{4}\right)\times \frac{49}{2}}}{2\left(-\frac{1}{4}\right)}$$

Multiply -4 times -\frac{1}{4}.

$$v_{2}=\frac{0±\sqrt{\frac{49}{2}}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of \frac{49}{2}.

$$v_{2}=\frac{0±\frac{7\sqrt{2}}{2}}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$v_{2}=\frac{0±\frac{7\sqrt{2}}{2}}{-\frac{1}{2}}$$

Now solve the equation v_{2}=\frac{0±\frac{7\sqrt{2}}{2}}{-\frac{1}{2}} when ± is plus.

$$v_{2}=-7\sqrt{2}$$

Now solve the equation v_{2}=\frac{0±\frac{7\sqrt{2}}{2}}{-\frac{1}{2}} when ± is minus.

$$v_{2}=7\sqrt{2}$$

The equation is now solved.

$$v_{2}=-7\sqrt{2} v_{2}=7\sqrt{2}$$