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How Do You Solve a Quadratic Equation With Complex Solutions by Using the Quadratic Formula?

Use the Quadratic Formula to solve x2 - 4x + 5 = 0 for x.

Summary

  1. Since there is no coefficient in front of x2, that means the 'a' value is 1
  2. When we plug in for 'b' we're taking the negative of -4, which gives us a positive 4
  3. '±' means 'plus or minus', which means we could end up with two answers depending on if we add or subtract
  4. 'i' is the imaginary unit, and it is equal to the square root of -1

Notes

    1. The Quadratic Formula only works if the equation is set equal to 0!
    2. That means we need to have 0 on one side and all the other terms on the other side
    1. 'a' is the coefficient in front of the x2 term
    2. Since there is no coefficient in front of x2, that means the 'a' value is 1
    3. 'b' is the coefficient in front of the x term, which is -4
    4. Remember, the sign of the coefficient needs to stay with it!
    5. 'c' is the constant term, which is 5
    1. 'a' is the coefficient in front of the x2 term
    2. Since there is no coefficient in front of x2, that means the 'a' value is 1
    3. 'b' is the coefficient in front of the x term, which is -4
    4. Remember, the sign of the coefficient needs to stay with it!
    5. 'c' is the constant term, which is 5
    1. Now that we know what a, b, and c are, we can plug those values into the Quadratic Formula!
    1. The symbol above the 'b2-4ac' is the symbol for square root
    2. '±' means 'plus or minus'
    3. Now that we know what a, b, and c are, we can plug those values into the Quadratic Formula!
    4. The a, b, and c values in the Quadratic Formula are the same as the ones we found in our quadratic equation
    5. So we can plug 1 in for 'a', -4 in for 'b', and 5 in for 'c'
    1. First simplify underneath the square root
    2. (-4)2 is 16, and -4•1•5 is -20
    3. Subtracting 16-20 gives us a negative, -4, under the square root
    1. 'Radical' is another term for the square root symbol
    2. Since we have the square root of a negative, we will have an 'i' in our answer
    3. Remember, 'i' is the imaginary unit, and it is equal to the square root of -1
    4. We can split up our radical into √(4)•√(-1)
    5. The square root of 4 is 2, and the square root of -1 is 'i'
    6. So √(-4) simplifies to 2i
    7. Dividing the top and bottom by 2 gives us a simplifed answer of 2±i
    8. This means our two solutions are 2+i and 2-i
    1. We need to plug 2+i and 2-i into the original equation to make sure we get true statements
    2. If we do, then we know our answers are correct!
    1. Plug '2+i' in anywhere you see an 'x'
    2. We can square complex numbers just like real numbers - just multiply 2+i times itself
    3. Distribute the -4 into the parentheses to get -8-4i
    4. FOILing together 2+i and 2+i gives us 4 for the First terms, 2i for the Outer terms, 2i for the Inner terms, and i2 for the Last terms
    5. i2 is the same as -1
    6. Terms with 'i' in them are like terms, just as if 'i' was a variable
    7. Our original equation was equal to 0, so 2+i is a correct solution!
    1. Plug '2-i' in anywhere you see an 'x'
    2. We can square complex numbers just like real numbers - just multiply 2-i times itself
    3. Distribute the -4 into the parentheses to get -8+4i
    4. FOILing together 2-i and 2-i gives us 4 for the First terms, -2i for the Outer terms, -2i for the Inner terms, and i2 for the Last terms
    5. i2 is the same as -1
    6. Terms with 'i' in them are like terms, just as if 'i' was a variable
    7. Our original equation was equal to 0, so 2-i is a correct solution!