Solve 4x+7sqrt{x}=2

$$4x+7 \sqrt{ x } =2$$

Solve for x

$$x=\frac{1}{16}=0.0625$$

Solution Steps

Subtract 4x from both sides of the equation.

$$7\sqrt{x}=2-4x$$

Square both sides of the equation.

$$\left(7\sqrt{x}\right)^{2}=\left(2-4x\right)^{2}$$

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

$$7^{2}\left(\sqrt{x}\right)^{2}=\left(2-4x\right)^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49\left(\sqrt{x}\right)^{2}=\left(2-4x\right)^{2}$$

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-4x\right)^{2}.

$$49x=4-16x+16x^{2}$$

Subtract 4 from both sides.

$$49x-4=-16x+16x^{2}$$

Add 16x to both sides.

$$49x-4+16x=16x^{2}$$

Combine 49x and 16x to get 65x.

$$65x-4=16x^{2}$$

Subtract 16x^{2} from both sides.

$$65x-4-16x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-16x^{2}+65x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=65 ab=-16\left(-4\right)=64$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 64.

$$1,64 2,32 4,16 8,8$$

Calculate the sum for each pair.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

The solution is the pair that gives sum 65.

$$a=64 b=1$$

Rewrite -16x^{2}+65x-4 as \left(-16x^{2}+64x\right)+\left(x-4\right).

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

Factor out 16x in the first and -1 in the second group.

$$16x\left(-x+4\right)-\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(16x-1\right)$$

To find equation solutions, solve -x+4=0 and 16x-1=0.

$$x=4 x=\frac{1}{16}$$

Substitute 4 for x in the equation 4x+7\sqrt{x}=2.

$$4\times 4+7\sqrt{4}=2$$

Simplify. The value x=4 does not satisfy the equation.

$$30=2$$

Substitute \frac{1}{16} for x in the equation 4x+7\sqrt{x}=2.

$$4\times \frac{1}{16}+7\sqrt{\frac{1}{16}}=2$$

Simplify. The value x=\frac{1}{16} satisfies the equation.

$$2=2$$

Equation 7\sqrt{x}=2-4x has a unique solution.

$$x=\frac{1}{16}$$