Solve 7^2+0.5x^2=4x^2

$${ 7 }^{ 2 } +0.5 { x }^{ 2 } =4 { x }^{ 2 }$$

Solve for x

$$x=\sqrt{14}\approx 3.741657387$$

$$x=-\sqrt{14}\approx -3.741657387$$

Steps by Finding Square Root

Calculate 7 to the power of 2 and get 49.

$$49+0.5x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$49+0.5x^{2}-4x^{2}=0$$

Combine 0.5x^{2} and -4x^{2} to get -3.5x^{2}.

$$49-3.5x^{2}=0$$

Subtract 49 from both sides. Anything subtracted from zero gives its negation.

$$-3.5x^{2}=-49$$

Divide both sides by -3.5.

$$x^{2}=\frac{-49}{-3.5}$$

Expand \frac{-49}{-3.5} by multiplying both numerator and the denominator by 10.

$$x^{2}=\frac{-490}{-35}$$

Divide -490 by -35 to get 14.

$$x^{2}=14$$

Take the square root of both sides of the equation.

$$x=\sqrt{14} x=-\sqrt{14}$$

Steps Using the Quadratic Formula

Calculate 7 to the power of 2 and get 49.

$$49+0.5x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$49+0.5x^{2}-4x^{2}=0$$

Combine 0.5x^{2} and -4x^{2} to get -3.5x^{2}.

$$49-3.5x^{2}=0$$

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.

$$-3.5x^{2}+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3.5 for a, 0 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-3.5\right)\times 49}}{2\left(-3.5\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-3.5\right)\times 49}}{2\left(-3.5\right)}$$

Multiply -4 times -3.5.

$$x=\frac{0±\sqrt{14\times 49}}{2\left(-3.5\right)}$$

Multiply 14 times 49.

$$x=\frac{0±\sqrt{686}}{2\left(-3.5\right)}$$

Take the square root of 686.

$$x=\frac{0±7\sqrt{14}}{2\left(-3.5\right)}$$

Multiply 2 times -3.5.

$$x=\frac{0±7\sqrt{14}}{-7}$$

Now solve the equation x=\frac{0±7\sqrt{14}}{-7} when ± is plus.

$$x=-\sqrt{14}$$

Now solve the equation x=\frac{0±7\sqrt{14}}{-7} when ± is minus.

$$x=\sqrt{14}$$

The equation is now solved.

$$x=-\sqrt{14} x=\sqrt{14}$$