Solve 4x+3/2+3,5/4x+3=4

$$\frac { 4 x + 3 } { 2 } + \frac { 3,5 } { 4 x + 3 } = 4$$

Solve for x

$$x=1$$

$$x=-0,5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{3}{4} since division by zero is not defined. Multiply both sides of the equation by 2\left(4x+3\right), the least common multiple of 2;4x+3.

$$\left(4x+3\right)\left(4x+3\right)+2\times 3,5=8\left(4x+3\right)$$

Multiply 4x+3 and 4x+3 to get \left(4x+3\right)^{2}.

$$\left(4x+3\right)^{2}+2\times 3,5=8\left(4x+3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9+2\times 3,5=8\left(4x+3\right)$$

Multiply 2 and 3,5 to get 7.

$$16x^{2}+24x+9+7=8\left(4x+3\right)$$

Add 9 and 7 to get 16.

$$16x^{2}+24x+16=8\left(4x+3\right)$$

Use the distributive property to multiply 8 by 4x+3.

$$16x^{2}+24x+16=32x+24$$

Subtract 32x from both sides.

$$16x^{2}+24x+16-32x=24$$

Combine 24x and -32x to get -8x.

$$16x^{2}-8x+16=24$$

Subtract 24 from both sides.

$$16x^{2}-8x+16-24=0$$

Subtract 24 from 16 to get -8.

$$16x^{2}-8x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -8 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 16\left(-8\right)}}{2\times 16}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 16\left(-8\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-8\right)±\sqrt{64-64\left(-8\right)}}{2\times 16}$$

Multiply -64 times -8.

$$x=\frac{-\left(-8\right)±\sqrt{64+512}}{2\times 16}$$

Add 64 to 512.

$$x=\frac{-\left(-8\right)±\sqrt{576}}{2\times 16}$$

Take the square root of 576.

$$x=\frac{-\left(-8\right)±24}{2\times 16}$$

The opposite of -8 is 8.

$$x=\frac{8±24}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{8±24}{32}$$

Now solve the equation x=\frac{8±24}{32} when ± is plus. Add 8 to 24.

$$x=\frac{32}{32}$$

Divide 32 by 32.

$$x=1$$

Now solve the equation x=\frac{8±24}{32} when ± is minus. Subtract 24 from 8.

$$x=-\frac{16}{32}$$

Reduce the fraction \frac{-16}{32} to lowest terms by extracting and canceling out 16.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=1 x=-\frac{1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{3}{4} since division by zero is not defined. Multiply both sides of the equation by 2\left(4x+3\right), the least common multiple of 2;4x+3.

$$\left(4x+3\right)\left(4x+3\right)+2\times 3,5=8\left(4x+3\right)$$

Multiply 4x+3 and 4x+3 to get \left(4x+3\right)^{2}.

$$\left(4x+3\right)^{2}+2\times 3,5=8\left(4x+3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9+2\times 3,5=8\left(4x+3\right)$$

Multiply 2 and 3,5 to get 7.

$$16x^{2}+24x+9+7=8\left(4x+3\right)$$

Add 9 and 7 to get 16.

$$16x^{2}+24x+16=8\left(4x+3\right)$$

Use the distributive property to multiply 8 by 4x+3.

$$16x^{2}+24x+16=32x+24$$

Subtract 32x from both sides.

$$16x^{2}+24x+16-32x=24$$

Combine 24x and -32x to get -8x.

$$16x^{2}-8x+16=24$$

Subtract 16 from both sides.

$$16x^{2}-8x=24-16$$

Subtract 16 from 24 to get 8.

$$16x^{2}-8x=8$$

Divide both sides by 16.

$$\frac{16x^{2}-8x}{16}=\frac{8}{16}$$

Dividing by 16 undoes the multiplication by 16.

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

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{1}{2}x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{1}{2}x=\frac{1}{2}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{3}{4} x-\frac{1}{4}=-\frac{3}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=1 x=-\frac{1}{2}$$