Question

Determine todos los Valores de a para que la ecuación sea 3ax^2 4ax+a+1=0 tenga una sola solución

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Answer 1

3ax² + 4ax + a + 1 = 0

d = discriminante

d = (4a)² - 4 × 3a × (a + 1)

d = 4a² - 12a

4a² -12a > 0

04a² -12a = 0

04a² -12a < 0

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{a \: \epsilon}}{\hspace{0pt}\above 2.2pt}} \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{a = 5}}{\hspace{0pt}\above 2.2pt}} \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{a \: \epsilon}}{\hspace{0pt}\above 2.2pt}}$

$\pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{a \: \epsilon}}{\hspace{0pt}\above 2.2pt}} \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{a = 5 \: (1 \: solucion \: real)}}{\hspace{0pt}\above 2.2pt}} \\ \pink{{\hspace{0 pt}\above 2.2pt}\boldsymbol{\mathsf{a \: \epsilon}}{\hspace{0pt}\above 2.2pt}}$

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