Solve sqrt{9x+7}-sqrt{x}-sqrt{16x-7}=0

$$\sqrt{ 9x+7 } - \sqrt{ x } - \sqrt{ 16x-7 } = 0$$

Solve for x

$$x=1$$

Solution Steps

Subtract -\sqrt{x}-\sqrt{16x-7} from both sides of the equation.

$$\sqrt{9x+7}=-\left(-\sqrt{x}-\sqrt{16x-7}\right)$$

To find the opposite of -\sqrt{x}-\sqrt{16x-7}, find the opposite of each term.

$$\sqrt{9x+7}=-\left(-\sqrt{x}\right)-\left(-\sqrt{16x-7}\right)$$

The opposite of -\sqrt{x} is \sqrt{x}.

$$\sqrt{9x+7}=\sqrt{x}-\left(-\sqrt{16x-7}\right)$$

The opposite of -\sqrt{16x-7} is \sqrt{16x-7}.

$$\sqrt{9x+7}=\sqrt{x}+\sqrt{16x-7}$$

Square both sides of the equation.

$$\left(\sqrt{9x+7}\right)^{2}=\left(\sqrt{x}+\sqrt{16x-7}\right)^{2}$$

Calculate \sqrt{9x+7} to the power of 2 and get 9x+7.

$$9x+7=\left(\sqrt{x}+\sqrt{16x-7}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x}+\sqrt{16x-7}\right)^{2}.

$$9x+7=\left(\sqrt{x}\right)^{2}+2\sqrt{x}\sqrt{16x-7}+\left(\sqrt{16x-7}\right)^{2}$$

Calculate \sqrt{x} to the power of 2 and get x.

$$9x+7=x+2\sqrt{x}\sqrt{16x-7}+\left(\sqrt{16x-7}\right)^{2}$$

Calculate \sqrt{16x-7} to the power of 2 and get 16x-7.

$$9x+7=x+2\sqrt{x}\sqrt{16x-7}+16x-7$$

Combine x and 16x to get 17x.

$$9x+7=17x+2\sqrt{x}\sqrt{16x-7}-7$$

Subtract 17x-7 from both sides of the equation.

$$9x+7-\left(17x-7\right)=2\sqrt{x}\sqrt{16x-7}$$

To find the opposite of 17x-7, find the opposite of each term.

$$9x+7-17x+7=2\sqrt{x}\sqrt{16x-7}$$

Combine 9x and -17x to get -8x.

$$-8x+7+7=2\sqrt{x}\sqrt{16x-7}$$

Add 7 and 7 to get 14.

$$-8x+14=2\sqrt{x}\sqrt{16x-7}$$

Square both sides of the equation.

$$\left(-8x+14\right)^{2}=\left(2\sqrt{x}\sqrt{16x-7}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-8x+14\right)^{2}.

$$64x^{2}-224x+196=\left(2\sqrt{x}\sqrt{16x-7}\right)^{2}$$

Expand \left(2\sqrt{x}\sqrt{16x-7}\right)^{2}.

$$64x^{2}-224x+196=2^{2}\left(\sqrt{x}\right)^{2}\left(\sqrt{16x-7}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$64x^{2}-224x+196=4\left(\sqrt{x}\right)^{2}\left(\sqrt{16x-7}\right)^{2}$$

Calculate \sqrt{x} to the power of 2 and get x.

$$64x^{2}-224x+196=4x\left(\sqrt{16x-7}\right)^{2}$$

Calculate \sqrt{16x-7} to the power of 2 and get 16x-7.

$$64x^{2}-224x+196=4x\left(16x-7\right)$$

Use the distributive property to multiply 4x by 16x-7.

$$64x^{2}-224x+196=64x^{2}-28x$$

Subtract 64x^{2} from both sides.

$$64x^{2}-224x+196-64x^{2}=-28x$$

Combine 64x^{2} and -64x^{2} to get 0.

$$-224x+196=-28x$$

Add 28x to both sides.

$$-224x+196+28x=0$$

Combine -224x and 28x to get -196x.

$$-196x+196=0$$

Subtract 196 from both sides. Anything subtracted from zero gives its negation.

$$-196x=-196$$

Divide both sides by -196.

$$x=\frac{-196}{-196}$$

Divide -196 by -196 to get 1.

$$x=1$$

Substitute 1 for x in the equation \sqrt{9x+7}-\sqrt{x}-\sqrt{16x-7}=0.

$$\sqrt{9\times 1+7}-\sqrt{1}-\sqrt{16\times 1-7}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Equation \sqrt{9x+7}=\sqrt{16x-7}+\sqrt{x} has a unique solution.

$$x=1$$