What are the solutions to the equation solvefor x,x^2-(t-1)x+t-2=0 ?

The solutions to the equation solvefor x,x^2-(t-1)x+t-2=0 are x=t-2,x=1

Find the zeros of solvefor x,x^2-(t-1)x+t-2=0

The zeros of solvefor x,x^2-(t-1)x+t-2=0 are x=t-2,x=1

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solve for

x, x^{2} - (t - 1) x + t - 2 = 0

x = t - 2, x = 1

Solution steps

x^{2} - (t - 1) x + t - 2 = 0

Expand

x^{2} - (t - 1) x + t - 2 : x^{2} - x t + x + t - 2

x^{2} - x t + x + t - 2 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + (- t + 1) x + t - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - t + 1 ) \pm ( - t + 1 ) ^{2} - 4 \cdot 1 \cdot ( t - 2 )}{2 \cdot 1}

Simplify

(- t + 1)^{2} - 4 \cdot 1 \cdot (t - 2) : t - 3

x_{1, 2} = \frac{- ( - t + 1 ) \pm ( t - 3 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - t + 1 ) + t - 3}{2 \cdot 1}, x_{2} = \frac{- ( - t + 1 ) - ( t - 3 )}{2 \cdot 1}

x = \frac{- ( - t + 1 ) + t - 3}{2 \cdot 1} : t - 2

x = \frac{- ( - t + 1 ) - ( t - 3 )}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = t - 2, x = 1

x^{2} - (7 + 3) \cdot x + 6 (1 + 3) = 0

a, (1 - m)^{2} (m^{2} + a^{2}) = 0

x^{2} + (m + n) \cdot x - 2 (m - n) = 0

a (x) = \frac{3}{16} x^{2} - 18 x + 670

x, x^{2} + 4 y^{2} + 9 z^{2} - 2 x - 4 y - 6 z + 3 = 0

What are the solutions to the equation solvefor x,x^2-(t-1)x+t-2=0 ?
The solutions to the equation solvefor x,x^2-(t-1)x+t-2=0 are x=t-2,x=1
Find the zeros of solvefor x,x^2-(t-1)x+t-2=0
The zeros of solvefor x,x^2-(t-1)x+t-2=0 are x=t-2,x=1