Question

Să se rezolve ecuațiile știind că admit soluțiile specificate:
$a) \: 3 {x}^{2} + ( {m}^{2} - 3m)x - 4 = 0 \: \\ x1 =- 2$
$b)( {m}^{2} + 6) {x}^{2} - (3m - 1)x + m - 2 = 0 \\ x1 = - 1$
$d)(m + 2) {x}^{2} + 3x - {m}^{2} = 0 \\ x1 = m$
$c)( {m}^{2} - 5) {x}^{2} + (3m - 2)x + {m}^{2} - 13 = 0 \\ x1 = 2$


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Answer 1

Explicație pas cu pas:

a)

$3 {( - 2)}^{2} + ( {m}^{2} - 3m)( - 2) - 4 = 0 \\ 12 - 2 {m}^{2} + 6m - 4 = 0 \\ - 2 {m}^{2} + 6m + 8 = 0 \\ {m}^{2} - 3m - 4 = 0 \\ (m + 1)(m - 4) = 0 \\ = > \\ m = - 1; \: m = 4$

$3x^{2}+4x-4=0$

$x_{1}x_{2}=-\frac{4}{3}=>x_{2}=\frac{2}{3}$

b)

$( {m}^{2} + 6) {( - 1)}^{2} - (3m - 1)( - 1) + m - 2 = 0 \\ {m}^{2} + 6 + 3m - 1 + m - 2 = 0 \\ {m}^{2} + 4m + 3 = 0 \\ (m + 3)(m + 1) = 0 \\ = > \\ m = - 3; \: m = - 1$

$m = - 3=>3x^{2}+2x-1=0$

$=>x_{1}x_{2}=-\frac{1}{3}=>x_{2}=\frac{1}{3}$

$m = - 1=>7x^{2}+4x-3=0$

$=>x_{1}x_{2}=-\frac{3}{7}=>x_{2}=\frac{3}{7}$

c)

$({m}^{2} - 5) \times {2}^{2} + (3m - 2) \times 2 + {m}^{2} - 13 = 0 \\ 4{m}^{2} - 20 + 6m - 4 + {m}^{2} - 13 = 0 \\ 5{m}^{2} + 6m - 37 = 0 \\ D = 36 + 4 \times 5 \times 37 = 36 + 740 = 776 > 0 \\ m_{1} = \frac{ - 6 - 2 \sqrt{194} }{10} = \frac{ - 3 - \sqrt{194} }{5} \\ m_{2} = \frac{ - 6 + 2 \sqrt{194} }{10} = \frac{ - 3 + \sqrt{194} }{5}$

$x_{1}+x_{2}=-\frac{3m-2}{m^{2}-5}=>x_{2}=\frac{55 - 4\sqrt{194} }{6};x_{2}=\frac{55 + 4\sqrt{194} }{6}$

d)

$(m + 2) {m}^{2} + 3m - {m}^{2} = 0 \\ m^{3} + 2 {m}^{2} + 3m - {m}^{2} = 0 \\ m^{3} + {m}^{2} + 3m = 0 \\ m( {m}^{2} + m + 3) = 0$

$m = 0$

${m}^{2} + m + 3 = 0 \\ D = 1 - 12 = - 11 < 0 \\ fara \: solutii \: reale$

$m = 0=>2x^{2}+3x=0$

$x_{1}=0\\x_{1}+x_{2}=-\frac{3}{m+2}=-\frac{3}{2}=>x_{2}=-\frac{3}{2}$