What is 6x^2+7x-3=0,x_{1}=x_{2} ?

The solution to 6x^2+7x-3=0,x_{1}=x_{2} is

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6x^2+7x-3=0,x_{1}=x_{2}

6 x^{2} + 7 x - 3 = 0, x_{1} = x_{2}

(x = \frac{1}{3}, x = - \frac{3}{2}, x_{1} \in R, x_{1} \in R, x_{2} = x_{1} x_{2} = x_{1})

Solution steps

[6 x^{2} + 7 x - 3 = 0 x_{1} = x_{2}]

Solve

6 x^{2} + 7 x - 3 = 0 : x = \frac{1}{3}, x = - \frac{3}{2}

Solve

x_{1} = x_{2}

for

x = \frac{1}{3} : (x = \frac{1}{3}, x_{1} \in R, x_{2} = x_{1})

Solve

x_{1} = x_{2}

for

x = - \frac{3}{2} : (x = - \frac{3}{2}, x_{1} \in R, x_{2} = x_{1})

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

6 x^{2} + 7 x - 3 = 0, x_{1} = x_{2}

are

(x = \frac{1}{3}, x = - \frac{3}{2}, x_{1} \in R, x_{1} \in R, x_{2} = x_{1} x_{2} = x_{1})

x - y = 2, 2 x^{2} + 9 y^{2} = 23

a + b = 4, a^{2} + b^{2} = 2

\frac{1}{2} x + \frac{1}{3} y = 2, \frac{5}{x} - 2 y = \frac{17}{3}

x + 3 y = 1, 3 x^{2} + 3 y^{2} = 43

u + 6 = 3, v - 4 = - 2 u + v