Solve 4x^2x+6x^2+3x+63=0

$$4 { x }^{ 2 } x+6 { x }^{ 2 } +3x+63=0$$

Solve for x (complex solution)

$$x=\frac{-5\sqrt{3}i+3}{4}\approx 0.75-2.165063509i$$

$$x=-3$$

$$x=\frac{3+5\sqrt{3}i}{4}\approx 0.75+2.165063509i$$

Solution Steps

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

$$6x^{2}+4xx^{2}+3x+63=0$$

Multiply and combine like terms.

$$4x^{3}+6x^{2}+3x+63=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 63 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{63}{4},±\frac{63}{2},±63,±\frac{21}{4},±\frac{21}{2},±21,±\frac{9}{4},±\frac{9}{2},±9,±\frac{7}{4},±\frac{7}{2},±7,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+6x^{2}+3x+63 by x+3 to get 4x^{2}-6x+21. Solve the equation where the result equals to 0.

$$4x^{2}-6x+21=0$$

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 4 for a, -6 for b, and 21 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4\times 21}}{2\times 4}$$

Do the calculations.

$$x=\frac{6±\sqrt{-300}}{8}$$

Solve the equation 4x^{2}-6x+21=0 when ± is plus and when ± is minus.

$$x=\frac{-5i\sqrt{3}+3}{4} x=\frac{3+5i\sqrt{3}}{4}$$

List all found solutions.

$$x=-3 x=\frac{-5i\sqrt{3}+3}{4} x=\frac{3+5i\sqrt{3}}{4}$$

Solve for x

$$x=-3$$

Solution Steps

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

$$6x^{2}+4xx^{2}+3x+63=0$$

Multiply and combine like terms.

$$4x^{3}+6x^{2}+3x+63=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 63 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{63}{4},±\frac{63}{2},±63,±\frac{21}{4},±\frac{21}{2},±21,±\frac{9}{4},±\frac{9}{2},±9,±\frac{7}{4},±\frac{7}{2},±7,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+6x^{2}+3x+63 by x+3 to get 4x^{2}-6x+21. Solve the equation where the result equals to 0.

$$4x^{2}-6x+21=0$$

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 4 for a, -6 for b, and 21 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4\times 21}}{2\times 4}$$

Do the calculations.

$$x=\frac{6±\sqrt{-300}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-3$$