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Sagot :
We are provided with a quadratic equation , and we have to find it's roots , but let's recall some things
- The Standard Form of a quadratic equation is ax² + bx + c = 0
- If D > 0 , then the equation have two real and distinct roots/zeroes .
- If D = 0 , then the equation have two equal roots/zeroes .
- If D < 0 , then the equation have two imaginary roots/zeroes or you can say that no real roots
Where , D is the Discriminant whose value is b²- 4ac and the roots of the equation are given by a very famous formula called The Quadratic Formula , which is as follows ;
[tex]{\quad {\boxed{\pmb{\bf x = \dfrac{-b \pm \sqrt{D}}{2a}}}}}[/tex]
Now , coming back on the question ;
We are provided with the equation x² + 6x = -18 , so on comparing this equation with the standard form of quadratic equation , we have , a = 1 , b = 6 and c = 18 . Now calculating the Discriminant ;
[tex]{: \implies \quad \sf D=(6)^{2}-4\times 1 \times 18}[/tex]
[tex]{: \implies \quad \sf D=36-72}[/tex]
[tex]{: \implies \quad \sf D=-18}[/tex]
As , D < 0 , so two imaginary roots exist . Now by quadratic formula ;
[tex]{: \implies \quad \sf x=\dfrac{-6 \pm \sqrt{-36}}{2\times 1}}[/tex]
[tex]{: \implies \quad \sf x=\dfrac{-6 \pm \sqrt{\underline{6\times 6}\times -1}}{2}}[/tex]
[tex]{: \implies \quad \sf x=\dfrac{-6 \pm 6\sqrt{-1}}{2}}[/tex]
[tex]{: \implies \quad \sf x=\dfrac{-6 \pm 6\iota}{2} \quad \{\because \sqrt{-1}=\iota\}}[/tex]
[tex]{: \implies \quad \sf x=\dfrac{2(-3\pm 3\iota)}{2}}[/tex]
[tex]{\quad \qquad \boxed{\bf \therefore \: x=-3 \pm 3\iota}}[/tex]
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