Find the slope of the line

Algebra Level 3

What is the sum of all possible solutions for x of the equation x ( x k ) = k + 1 x ( x - k ) = k + 1 ?

k 1 0 2k - 1 k+1

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3 solutions

Zach Abueg
Jun 16, 2017

By Vieta's formulas, the sum of the roots x 1 + x 2 \displaystyle x_1 + x_2 of a quadratic equation a x 2 + b x + c \displaystyle ax^2 + bx + c is b a \displaystyle -\dfrac ba .

x ( x k ) = k + 1 x 2 k x ( k + 1 ) = 0 b a = k 1 = k \displaystyle \begin{aligned} x(x - k) & = k + 1 \\ x^2 - kx - (k + 1) & = 0 \\ \implies -\dfrac ba & = -\dfrac{- k}{1} = k \end{aligned}

Engin Akpinar
Jun 16, 2017

Any quadratic equations in x with two roots a and b can be expressed as, ( x a ) ( x b ) = 0 (x - a)(x - b) = 0 . Which on expansion takes the form, x ² ( a + b ) x a b = 0 x² - (a + b)x - ab = 0 Hence, the sum of the roots = -(coefficient of x in the expanded form)

Now, x ( x k ) = k + 1 x(x - k) = k + 1 => x ² k x ( k + 1 ) = 0 x² - kx - (k + 1) = 0

Hence, sum of the roots = ( k ) = k = -(-k) = k

Toby M
Sep 25, 2020

You can actually factorise:

x 2 k x ( k + 1 ) = 0 x^2 - kx - (k+1) = 0 ( x ( k + 1 ) ) ( x + 1 ) = 0 \Rightarrow \left(x-(k+1) \right) \left(x+1 \right)= 0 x = k + 1 , 1 \Rightarrow x = k + 1, - 1

and the sum of roots is just k + 1 1 = k k + 1 - 1 = \boxed{k} .

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