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pashok25 [27]
3 years ago
5

Solving quadratic equations with complex solutions solve x^2-3x=-8

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
4 0

Answer:

\displaystyle  \large{ x =   \frac{  3 \pm \sqrt{ 23}i  }{2} }

Step-by-step explanation:

We are given the equation:-

\displaystyle  \large{ {x}^{2}  - 3x =  - 8}

First, arrange the expression to the standard form.

Recall the standard form of Quadratic Equation:-

\displaystyle  \large{a {x}^{2}   +  bx  + c = 0}

Therefore, add both sides by 8.

\displaystyle  \large{ {x}^{2}  - 3x  + 8=  - 8 + 8} \\  \displaystyle  \large{ {x}^{2}  - 3x  + 8=  0}

Since we know that the solutions are complex. Therefore, we apply our Quadratic Formula.

<u>Q</u><u>u</u><u>a</u><u>d</u><u>r</u><u>a</u><u>t</u><u>i</u><u>c</u><u> </u><u>F</u><u>o</u><u>r</u><u>m</u><u>u</u><u>l</u><u>a</u>

\displaystyle  \large{ x =   \frac{  - b \pm  \sqrt{ {b}^{2}  - 4ac} }{2a} }

From the equation, the value of:

  • a = 1
  • b = -3
  • c = 8

Substitute these values in the formula.

\displaystyle  \large{ x =   \frac{  - ( - 3)\pm  \sqrt{ {( - 3)}^{2}  - 4(1)(8)} }{2(1)} } \\  \displaystyle  \large{ x =   \frac{  3\pm  \sqrt{ 9  -32} }{2} } \\  \displaystyle  \large{ x =   \frac{  3 \pm \sqrt{ - 23} }{2} }

Make sure to recall every necesscary and important fundamental math such as multiplying with negative, exponents, etc.

<u>I</u><u>m</u><u>a</u><u>g</u><u>i</u><u>n</u><u>a</u><u>r</u><u>y</u><u> </u><u>U</u><u>n</u><u>i</u><u>t</u>

\displaystyle \large{i =  \sqrt{ - 1} } \\  \displaystyle \large{ {i}^{2}  =   - 1} \\  \displaystyle \large{ {i}^{3} = -   \sqrt{ - 1}  =  - i} \\  \displaystyle \large{ {i}^{4} =   {i}^{2}  \times  {i}^{2}   =   ( - 1)( - 1) = 1}

From √-23, factor √-1 out.

\displaystyle  \large{ x =   \frac{  3 \pm \sqrt{ 23} \sqrt{ - 1}  }{2} }

Convert √-1 to i.

\displaystyle  \large{ x =   \frac{  3 \pm \sqrt{ 23}i  }{2} }

And we're done!

abruzzese [7]3 years ago
3 0

x^2-3x=-8\\\\x^2-3x+2.25=-5.75\\\\(x-1.5)^2=-5.75\\x-1.5=\pm i\sqrt{5.75}\\\boxed{x=1.5 \pm i\sqrt{5.75}}

Not the prettiest but you can simplify

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