Question

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$(2m+3)+ \frac{12}{(2m+3)}+13=0$
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Answer 1

Answer:

m = -7.5 , -2

Step-by-step explanation:

 $\sf (2m+3)+\dfrac{12}{2m+3}+13=0 \\\\\\ \sf (2m+3)+\dfrac{12}{2m+3}=-13 \\\\\\ \sf \dfrac{(2m+3)(2m+3)}{2m+3}+\dfrac{12}{2m+3}=-13 \\\\\\ \sf \dfrac{(2m+3)^2}{2m+3}+\dfrac{12}{2m+3}=-13 \\\\\\ \sf \dfrac{(2m)^2+3^2+2(2m)(3)}{2m+3}+\dfrac{12}{2m+3}=-13 \\\\\\ \sf \dfrac{4m^2+9+12m}{2m+3}+\dfrac{12}{2m+3}=-13 \\\\\\ \sf \dfrac{4m^2+9+12m+12}{2m+3}=-13 \\\\\\ \sf \dfrac{4m^2+12m+21}{2m+3}=-13$

 4m² + 12m + 21 = -13(2m + 3)

  4m² + 12m + 21 = -26m - 39

  4m² + 12m + 26m + 21 + 39 = 0

  4m² + 38m + 60 = 0

  2(2m² + 19m + 30) = 0

  2m² + 19m + 30 = 0

Solving the quadratic equation by quadratic formula,

 $\boxed{\tt x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$

 $\implies \tt m=\dfrac{-19 \pm \sqrt{(-19)^2-4(2)(30)}}{2(2)} \\\\ \implies \tt m=\dfrac{-19 \pm \sqrt{361-240}}{4} \\\\ \implies \tt m=\dfrac{-19 \pm \sqrt{121}}{4} \\\\ \implies \tt m=\dfrac{-19 \pm 11}{4} \\\\ \implies \tt m=\dfrac{-19-11}{4} \ ; \ m=\dfrac{-19+11}{4} \\\\ \implies \tt m=\dfrac{-30}{4} \ ; \ m=\dfrac{-8}{4} \\\\ \implies \tt m=-7.5 \ ; \ m=-2$

Therefore, m = -7.5 , -2

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Identity used : (a + b)² = a² + b² + 2ab