Solve (4x+5)=16x^2-25/4x^2+23x-35

$$( 4 x + 5 ) = \frac { 16 x ^ { 2 } - 25 } { 4 x ^ { 2 } + 23 x - 35 }$$

Solve for x

$$x=-6$$

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Solution Steps

Variable x cannot be equal to any of the values -7,\frac{5}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-5\right)\left(x+7\right).

$$4x\left(4x-5\right)\left(x+7\right)+\left(4x-5\right)\left(x+7\right)\times 5=16x^{2}-25$$

Use the distributive property to multiply 4x by 4x-5.

$$\left(16x^{2}-20x\right)\left(x+7\right)+\left(4x-5\right)\left(x+7\right)\times 5=16x^{2}-25$$

Use the distributive property to multiply 16x^{2}-20x by x+7 and combine like terms.

$$16x^{3}+92x^{2}-140x+\left(4x-5\right)\left(x+7\right)\times 5=16x^{2}-25$$

Use the distributive property to multiply 4x-5 by x+7 and combine like terms.

$$16x^{3}+92x^{2}-140x+\left(4x^{2}+23x-35\right)\times 5=16x^{2}-25$$

Use the distributive property to multiply 4x^{2}+23x-35 by 5.

$$16x^{3}+92x^{2}-140x+20x^{2}+115x-175=16x^{2}-25$$

Combine 92x^{2} and 20x^{2} to get 112x^{2}.

$$16x^{3}+112x^{2}-140x+115x-175=16x^{2}-25$$

Combine -140x and 115x to get -25x.

$$16x^{3}+112x^{2}-25x-175=16x^{2}-25$$

Subtract 16x^{2} from both sides.

$$16x^{3}+112x^{2}-25x-175-16x^{2}=-25$$

Combine 112x^{2} and -16x^{2} to get 96x^{2}.

$$16x^{3}+96x^{2}-25x-175=-25$$

Add 25 to both sides.

$$16x^{3}+96x^{2}-25x-175+25=0$$

Add -175 and 25 to get -150.

$$16x^{3}+96x^{2}-25x-150=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -150 and q divides the leading coefficient 16. List all candidates \frac{p}{q}.

$$±\frac{75}{8},±\frac{75}{4},±\frac{75}{2},±75,±150,±\frac{75}{16},±\frac{25}{8},±\frac{25}{4},±\frac{25}{2},±25,±50,±\frac{15}{8},±\frac{15}{4},±\frac{15}{2},±15,±30,±\frac{25}{16},±\frac{15}{16},±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±10,±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±6,±\frac{5}{16},±\frac{3}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1,±2,±\frac{1}{16}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 16x^{3}+96x^{2}-25x-150 by x+6 to get 16x^{2}-25. Solve the equation where the result equals to 0.

$$16x^{2}-25=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 16 for a, 0 for b, and -25 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 16\left(-25\right)}}{2\times 16}$$

Do the calculations.

$$x=\frac{0±40}{32}$$

Solve the equation 16x^{2}-25=0 when ± is plus and when ± is minus.

$$x=-\frac{5}{4} x=\frac{5}{4}$$

Remove the values that the variable cannot be equal to.

$$x=-\frac{5}{4}\text{ or }x=-6$$

List all found solutions.

$$x=-6 x=-\frac{5}{4} x=\frac{5}{4}$$

Variable x cannot be equal to \frac{5}{4}.

$$x=-\frac{5}{4} x=-6$$