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The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3
Question 11
The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$. (a) Show that $k$ satisfies $k^2 - 2k - 24 < 0$ ... show full transcript
Worked Solution & Example Answer:The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3
Step 1
Show that k satisfies $k^2 - 2k - 24 < 0$
Answer
To show that the equation has two distinct real solutions, we need to analyze the quadratic equation formed.
Using the general form of a quadratic equation , we identify:
For the quadratic to have two distinct real solutions, the discriminant must be positive:
Substituting the values, we have:
Calculating gives: Expanding the product:
Setting leads to: Dividing by -4 and flipping the inequality: This shows that satisfies the given inequality.
Step 2
Hence find the set of possible values of k
Answer
Next, we solve the inequality . To find the critical points, we can factor the quadratic:
Setting this equal to zero gives critical points:
ightarrow k = 6$$ $$k + 4 = 0 ightarrow k = -4$$ The solution to the inequality can be found by testing intervals around the critical points $k = -4$ and $k = 6$: - For $k < -4$: Say $k = -5$, then $(-5 - 6)(-5 + 4) = (-11)(-1) > 0$ (not a solution). - For $-4 < k < 6$: Say $k = 0$, then $(0 - 6)(0 + 4) = (-6)(4) < 0$ (a solution). - For $k > 6$: Say $k = 7$, then $(7 - 6)(7 + 4) = (1)(11) > 0$ (not a solution). Thus, the set of possible values for $k$ is: $$-4 < k < 6$$