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Let f(x) = x^2 + (k + 3)x + k where k is a real constant - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 1

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Let---f(x)-=-x^2-+-(k-+-3)x-+-k---where-k-is-a-real-constant-Edexcel-A-Level Maths Pure-Question 9-2011-Paper 1.png

Let f(x) = x^2 + (k + 3)x + k where k is a real constant. (a) Find the discriminant of f(x) in terms of k. (b) Show that the discriminant of f(x) can b... show full transcript

Worked Solution & Example Answer:Let f(x) = x^2 + (k + 3)x + k where k is a real constant - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 1

Step 1

Find the discriminant of f(x) in terms of k.

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Answer

The discriminant of a quadratic function of the form ax2+bx+cax^2 + bx + c is given by the formula D=b24acD = b^2 - 4ac. For the function
f(x) = x^2 + (k + 3)x + k:\n\nHere, a = 1, b = k + 3, and c = k.\n\nSubstituting these values into the formula gives us:
D=(k+3)24(1)(k)D = (k + 3)^2 - 4(1)(k)
Expanding this results in:
D=(k2+6k+9)4kD = (k^2 + 6k + 9) - 4k This simplifies to:
D=k2+2k+9D = k^2 + 2k + 9

Step 2

Show that the discriminant of f(x) can be expressed in the form (k + a)^2 + b, where a and b are integers to be found.

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Answer

We have already derived the discriminant as
D=k2+2k+9.D = k^2 + 2k + 9.
We can manipulate this expression to fit the required form.
\nNotice that we can complete the square for the quadratic term:
D=(k2+2k+1)+8=(k+1)2+8.D = (k^2 + 2k + 1) + 8 = (k + 1)^2 + 8.
Thus, we can state that:
a = 1 and b = 8.

Step 3

Show that, for all values of k, the equation f(x) = 0 has real roots.

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Answer

In order for the quadratic equation f(x) = 0 to have real roots, the discriminant must be non-negative, i.e., D0D \geq 0.
We previously established that:
D=(k+1)2+8.D = (k + 1)^2 + 8.
Here, (k+1)2(k + 1)^2 is always non-negative since it is a square term, and adding 8 ensures that
$$D \geq 8 > 0.$
Therefore, for all values of k, the equation f(x) = 0 will always have real roots.

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