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Solve for x (complex solution)
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±15,±5,±3,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=3
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^{2}-4x+5=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+17x-15 by x-3 to get x^{2}-4x+5. Solve the equation where the result equals to 0.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 5}}{2}
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 1 for a, -4 for b, and 5 for c in the quadratic formula.
x=\frac{4±\sqrt{-4}}{2}
Do the calculations.
x=2-i x=2+i
Solve the equation x^{2}-4x+5=0 when ± is plus and when ± is minus.
x=3 x=2-i x=2+i
List all found solutions.
±15,±5,±3,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=3
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^{2}-4x+5=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+17x-15 by x-3 to get x^{2}-4x+5. Solve the equation where the result equals to 0.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 5}}{2}
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 1 for a, -4 for b, and 5 for c in the quadratic formula.
x=\frac{4±\sqrt{-4}}{2}
Do the calculations.
x\in \emptyset
Since the square root of a negative number is not defined in the real field, there are no solutions.
x=3
List all found solutions.