The solution of the quadratic equation [tex]6x^{2} + 6 = 12x + 18[/tex] are [tex]1 + \sqrt{3}[/tex] and [tex]1 - \sqrt{3}[/tex].
Roots of Quadratic Equations:
- For the general quadratic equation, [tex]ax^{2} + bx + c = 0[/tex], the roots are given by
[tex]x = \frac{ - b + \sqrt{D} }{2a}[/tex] or [tex]x = \frac{ - b - \sqrt{D} }{2a}[/tex]
where, D = Discriminant = [tex]b^{2} - 4 a c[/tex].
How to solve the given quadratic equation?
- Arrange the given equation in the general form.
∴ [tex]6x^{2} + 6 - 12x -18 = 0[/tex]
∴ [tex]6x^{2} - 12x -12 = 0[/tex] - Find the discriminant of the equation.
∴ comparing the coefficients from the general form,
a = 6 , b = -12 & c = -12
∴ Discriminant of the given equation,
∴ [tex]D = b^{2} - 4 a c = (-12)^2 - 4 (6) (-12) = 144 \ + \ 288 = 432[/tex] - Check the nature of roots.
As D > 0, the roots of the equation are real and distinct. - Find the roots.
∴ [tex]x = \frac{ - b + \sqrt{D} }{2a} = \frac{ - (-12) + \sqrt{432} }{2 \ (6)} = \frac{12 + 12 \sqrt{3} }{12} = 1 + \sqrt{3}[/tex] &
∴ [tex]x = \frac{ - b - \sqrt{D} }{2a} = \frac{ - (-12) - \sqrt{432} }{2 \ (6)} = \frac{12 - 12 \sqrt{3} }{12} = 1 - \sqrt{3}[/tex]
Thus the solutions of the given equations are [tex]1 + \sqrt{3}[/tex] and [tex]1 - \sqrt{3}[/tex].
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