smc-984-50-Discriminant

Functions, 2ADV F1 EQ-Bank 20

Find the value of \(k\) if \(4kx^2-(3-4k) x+k=0\) has one root. (2 marks)

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\(k=\dfrac{3}{8}\)

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\(4kx^2-(3-4k) x+k=0\)

\(\text{1 root}\ \Rightarrow \Delta=0\)

	\(\Delta\)	\(=b^2-4 a c\)
	\(0\)	\(=\left[ -\left( 3-4k \right)\right]^2-4\times 4k \times k\)
	\(0\)	\(=9-24k + 16k^2-16k^2\)
	\(24k\)	\(=9\)
	\(k\)	\(=\dfrac{9}{24}=\dfrac{3}{8}\)

The tangent to the parabola \(y=x^2+2 x-4\) is \(y=px-5\) where \(p>0\).

Find the value of \(p\). (2 marks)

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\(p=4\)

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\(\text{Intersection occurs when:}\)

\(x^2+2x-4\)	\(=px-5\)
\(x^2+(2-p)x+1\)	\(=0\)

\(\text{Tangent touches once}\ \Rightarrow\ \text{Discriminant}\ \Delta=0\)

COMMENT: Key is to recognise this is a discriminant question, not a calculus application.

Show that the parabola \(2x^2-kx+k-2\) has at least one real root. (3 marks)

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\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\)	\(\geqslant 0\)
\((x-4)^2\)	\(\geqslant 0\)

\(\therefore\ \text{At least one root exists for all}\ k\)

Show Worked Solution

\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\)	\(\geqslant 0\)
\((x-4)^2\)	\(\geqslant 0\)

\(\therefore\ \text{At least one root exists for all}\ k\)

Find the values of `k` for which the expression `x^2-3x + (4-2k)` is always positive. (3 marks)

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`k < 7/8`

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`x^2-3x + (4-2k) > 0`

`x^2-3x + (4-2k) = 0\ \ text(is a concave up parabola)`

`=>\ text{Always positive (no roots) if}\ \ Delta < 0`

`b^2-4ac < 0`