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Which of the following is a solution of x^2+6x=-18

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]