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The equation $kx^2 + 4x + (5 - k) = 0$, where $k$ is a constant, has 2 different real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 9 - 2009 - Paper 1
Question 9
The equation $kx^2 + 4x + (5 - k) = 0$, where $k$ is a constant, has 2 different real solutions for $x$. (a) Show that $k$ satisfies $$k^2 - 5k + 4 > 0.$$ (b... show full transcript
Worked Solution & Example Answer:The equation $kx^2 + 4x + (5 - k) = 0$, where $k$ is a constant, has 2 different real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 9 - 2009 - Paper 1
Step 1
Show that $k$ satisfies $k^2 - 5k + 4 > 0$
Answer
To show that satisfies the inequality , we need to analyze the discriminant of the quadratic equation. The quadratic formula gives us solutions for as:
ightarrow ext{ } ext{if }} { ext{1}} ext{ } > rac{(- ext{in this case, -5}) ± ext{√(5^2 - 4 * 1 * (5 - k))}} {2 * 1}$$ For there to be two distinct real solutions, the discriminant must be positive: $$b^2 - 4ac > 0 ightarrow (-5)^2 - 4 * 1 * (5 - k) > 0$$ This simplifies to: $$25 - 20 + 4k > 0$$ $$4k + 5 > 0$$ Thus, rearranging gives us the desired inequality: $$k^2 - 5k + 4 > 0$$.Step 2
Hence find the set of possible values of $k$
Answer
To find the set of possible values of , we will factor the quadratic inequality:
We need to analyze the intervals determined by the roots and . Since this is a quadratic opening upwards (as the coefficient of is positive), the quadratic will be greater than zero outside the interval defined by these roots.
Thus, the solutions to the inequality occur when:
Therefore, the set of possible values for is: