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If one of the root of the quadratic equation kx2 + (a+b) x + ab = 0 is – b, find the value of k

User Filnor
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1 Answer

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Final Answer:

The value of
\( k \) is \( 1 \).

Step-by-step explanation:

The quadratic equation given is
\( kx^2 + (a+b)x + ab = 0 \), and one of its roots is given as
\( -b \). In a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is
\( (-b)/(a) \), and the product of the roots is
\( (c)/(a) \). In this case, the sum of the roots is
\( (-(a+b))/(k) \) and the product is \( (ab)/(k) \). Since one of the roots is given as
\( -b \), the other root is \( (ab)/(k) \). The sum of the roots is therefore
\( -b + (ab)/(k) \), and according to the given information, this is equal to
\( -b \).

Setting up the equation:


\[ -b + (ab)/(k) = -b \]

Simplifying, we get
\( (ab)/(k) = 0 \), which implies \( ab = 0 \). Since
\( a \) and \( b \)are the coefficients in the quadratic equation, at least one of them must be zero. Assuming
\( b \) is non-zero, \( a \) must be zero. Therefore,
\( k \) must be \( 1 \) to satisfy the given conditions.

User Johannes Reichard
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