O lvl A Maths: Quadratic Equations

Question from http://www.sgforums.com/forums/2297/topics/401082

Given that y = (x² + 2x - c) / (x - 1) and x is real, find the range of values of c for which y can take all real values.

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Answer:

y = (x² + 2x - c) / (x - 1)

y (x - 1 ) = x² + 2x - c

yx - y = x² + 2x - c

x² + 2x - yx - c + y = 0

x² + (2 - y) x + (y - c) = 0

Since x is real, discriminant ≥ 0

(2 - y)² - 4 (1) (y - c) ≥ 0

4 - 4y + y² - 4y + 4c ≥ 0

y² - 8y + (4c + 4) ≥ 0

Since the equation needs to be ≥ 0, it means it has either equal roots or no real roots

so (-8)² - 4 (1) (4c + 4) ≤ 0
64 - 16c - 16 ≤ 0
48 - 16c ≤ 0
48 ≤ 16c
c ≥ 3 (ans)



Graph example for clarity:

Plotted with
(i) c = 2
(ii) c = 4
(iii) c = 3

O lvl A Maths: Quadratic Equations

Find the value of m if the roots of the quadratic equation x² + 2mx + m + 2 = 0 are equal

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Answer:

When the roots are equal, discriminant (b²-4ac) = 0

In this case, a = 1, b = 2m and c = m+2

b² - 4ac = 4m² - 4(1)(m+2) = 0
4m² - 4m - 8 = 0
m² - m - 2 = 0
(m-2)(m+1) = 0
m = 2 or -1

O lvl A Maths: Quadratic Equations

Questions from http://www.sgforums.com/forums/2297/topics/312370

Show that the solutions of the equation x² + kx = 3 - k are real for all real values of k.

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Answer:

Shift all to one side
x² + kx - 3 + k = 0
x² + kx + (k - 3) = 0

For all solutions to be real, we must show that the discriminant is ≥ 0

Discriminant = b² - 4ac
= k² - 4(1)(k - 3)
= k² - 4k + 12
= k² - 4k + 4 + 8
= (k - 2)² + 8

Since (k - 2)² ≥ 0, (k - 2)² + 8 ≥ 0
Hence discriminant is ≥ 0

Thus, the solutions of the equation x² + kx = 3 - k are real for all real values of k (shown).

O lvl A Maths: Quadratic Equations

Questions from http://www.sgforums.com/forums/2297/topics/311161

1) Show that x² + hx + k= 0 has two distinct roots for all negative values of k.

2) Find the values of k for which the x axis is a tangent to the curve y = 3x² - 8x+ 5 - k. For each value of k, find the coordinates for the point of tangency.

3) Find the range of values of k for which the equation x² + 5x + 3 = k has two real and distinct roots.

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Answer:

1)

For
x² + hx + k = 0

discriminant = h² - 4(1)(k) = h² -4k

h² ≥ 0
also, for negative values of k, -4k > 0

Therefore h² - 4k is always +ve, ie >0

Thus, since discriminant is always > 0, equation will always have 2 distinct roots.

2) x-axis tangent to the curve => curve touches the line y=0 only 1 time

so 3x² - 8x + 5 - k = 0 has only 1 root (real and equal roots).

Discriminant = 0 for real and equal roots
64 - 4(3)(5 - k) = 0
4 + 12k = 0
k = -⅓

Find value of x: 2 methods

1st method:
3x² - 8x + 16/3 = 0
9x² - 24x + 16 = 0
(3x - 4)² = 0
x = 4/3
and y = 0 (since it is on x-axis)

Thus
coordinates of tangential point: (4/3, 0)

2nd method:
Obviously, for an x² curve, ie curve with smiling face, the minimum point in this case will be the point at tangent with the x-axis.

so dy/dx = 6x-8
x = 4/3

Similarly, y = 0 (since it is on x-axis)

Thus
coordinates of tangential point: (4/3, 0)

O lvl A Maths: Quadratic Equations

a) The roots of the equation 3x² + kx + 96 =0 are both positive and one is twice as large as the other Calculate the value of each root and find k

b) Given that p² +q² = 13 and that pq = 6, construct the quadratic equation whose roots are p² and q². Find all possible values of p.

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Answer:

a)

Let a be the first root, and 2a be the second root (twice as large)

Therefore, (x-2a) (x-a) = 0 is the equation for the roots.

Expanding
x² - 3ax + 2a² = 0
3x² - 9ax + 6a² = 0

Comparing with 3x² + kx + 96 = 0,
6a² = 96

thus,
a is 4 or -4 (reject because a is positive)

So the value of the roots are 4 and 8 (a and 2a)

Solving for k, k = 9a = -36

b)

For a quadratic equation whose roots are p² and q²,
(x-p²)(x-q²) = 0
x² - (p²+q²)x + p²q² = 0
x² - 13x + 36 = 0

The equation can be factorised to
x² - 13x + 36 = 0 ⇒ (x - 4)(x - 9) = 0

thus,

p² = 4 or p² = 9
p = ±2 or ±3

O lvl A Maths: Quadratic Equations

Show that the quadratic equation k(x² - 2x) + 2 = 3x² - k has no real roots if k < -6.

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Answer:

k(x² - 2x) + 2 = 3x² - k
kx² - 2kx + 2 = 3x² - k
kx² - 2kx + 2 - 3x² + k = 0
(k - 3)x² - 2kx + (k + 2) = 0

For no real roots, b² - 4ac < 0

Hence,
(-2k)² - 4(k - 3)(k + 2) < 0
4k² - 4(k² - k - 6) < 0
k² - (k² - k - 6) < 0
k² - k² + k + 6 < 0
k < -6

O lvl A Maths: Quadratic Equations

Show that the quadratic equation 2k (x+1) = 3 - 3x² has real roots for all real values of k.

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Answer:

2k (x+1) = 3 - 3x²
2kx + 2k = 3 - 3x²
2kx + 2k - 3 + 3x² = 0
3x² + 2kx + (2k - 3) = 0

To show that an equation has only real roots, we must show that the discriminant ≥ 0
i.e. b² - 4ac ≥ 0

b² - 4ac = (2k)² - 4(3)(2k - 3)
= 4k² - 24k + 36
= 4(k² - 6k + 9)
= 4(k - 3)²

Since k is real, (k - 3)² ≥ 0

It follows that b² - 4ac ≥ 0

Thus, the equation has real roots for all real values of k (shown)