Quadratics

1. Question: Solve $3x^2 - 7x + 2 = 0$ by factoring.

$(3x - 1)(x - 2) = 0 \implies x = 1/3$ or $x = 2$.

2. Question: Find the values of $k$ for which $kx^2 + 2x + k = 0$ has two distinct real roots.

$\Delta = 4 - 4k^2 \gt 0 \implies k^2 \lt 1 \implies -1 \lt k \lt 1$ (and $k \neq 0$ since it must Be quadratic).

3. Question: Express $f(x) = -x^2 + 6x - 5$ in completed square form and state its maximum value.

$f(x) = -(x^2 - 6x) - 5 = -[(x - 3)^2 - 9] - 5 = -(x - 3)^2 + 4$. Maximum value: $4$ at $x = 3$.

4. Question: Solve the inequality $2x^2 + 3x - 5 \lt 0$.

$2x^2 + 3x - 5 = (2x + 5)(x - 1)$. Roots: $x = -5/2$ and $x = 1$. Parabola opens upward. Solution: $-5/2 \lt x \lt 1$.

5. Question: If $\alpha$ and $\beta$ are roots of $x^2 - 3x + 5 = 0$Form a quadratic equation with roots $\alpha + 2$ and $\beta + 2$.

New sum: $(\alpha + 2) + (\beta + 2) = (\alpha + \beta) + 4 = 3 + 4 = 7$.

New product: $(\alpha + 2)(\beta + 2) = \alpha\beta + 2(\alpha + \beta) + 4 = 5 + 6 + 4 = 15$.

Equation: $x^2 - 7x + 15 = 0$.

6. Question: A ball is thrown upward. Its height $h$ metres at time $t$ seconds is $h = -5t^2 + 20t + 1$. Find the maximum height and the time at which it occurs.

$h = -5(t^2 - 4t) + 1 = -5[(t - 2)^2 - 4] + 1 = -5(t - 2)^2 + 21$. Maximum height: $21 \mathrm{ m}$ At $t = 2 \mathrm{ s}$.

7. Question: Find the range of the function $f(x) = 2x^2 - 8x + 11$.

$f(x) = 2(x^2 - 4x) + 11 = 2[(x - 2)^2 - 4] + 11 = 2(x - 2)^2 + 3$. Since $2(x - 2)^2 \geqslant 0$ The minimum is $3$. Range: $[3, \infty)$.

8. Question: The equation $x^2 + 2px + p + 8 = 0$ has real roots. Find the range of $p$.

$\Delta = 4p^2 - 4(p + 8) = 4p^2 - 4p - 32 \geqslant 0 \implies p^2 - p - 8 \geqslant 0$.

Roots: $p = \dfrac{1 \pm \sqrt{1 + 32}}{2} = \dfrac{1 \pm \sqrt{33}}{2}$.

Since $a = 1 \gt 0$: $p \leqslant \dfrac{1 - \sqrt{33}}{2}$ or $p \geqslant \dfrac{1 + \sqrt{33}}{2}$.

9. Question: Prove that the quadratic equation $(a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0$ has no real roots unless $ad = bc$.

$\Delta = 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 4[a^2c^2 + 2abcd + b^2d^2 - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2]$ $= 4[2abcd - a^2d^2 - b^2c^2] = -4(ad - bc)^2$.

Since $-(ad - bc)^2 \leqslant 0$We have $\Delta \leqslant 0$. Real roots exist only when $\Delta = 0$I.e., $ad = bc$.

10. Question: For what integer values of $k$ does the equation $x^2 - kx + k + 3 = 0$ have integer roots?

By Vieta: $\alpha + \beta = k$ and $\alpha\beta = k + 3$. So $\alpha\beta = \alpha + \beta + 3$ Giving $\alpha\beta - \alpha - \beta = 3$Or $(\alpha - 1)(\beta - 1) = 4$.

Factor pairs of 4: $(1,4)$, $(2,2)$, (4,1)$$(-1,-4)$$(-2,-2)$$(-4,-1).

These give (\alpha, \beta) = (2,5)$$(3,3)$$(5,2)$$(0,-3)$$(-1,-1)$$(-3,0).

Corresponding $k$ values: $7, 6, 7, -3, -2, -3$. Integer values of $k$: $\{-3, -2, 6, 7\}$.

Case 1: $k = 1$. The equation becomes $2x + 4 = 0 \implies x = -2$. One real root, so $k = 1$ is included.

Case 2: $k \neq 1$. This is a genuine quadratic. Require $\Delta \geq 0$:

$\Delta = (2k)^2 - 4(k-1)(k+3) = 4k^2 - 4(k^2 + 2k - 3) = 4k^2 - 4k^2 - 8k + 12 = -8k + 12$

$-8k + 12 \geq 0 \implies k \leq \frac{3}{2}$

Combined with Case 1: the answer is $k \leq \dfrac{3}{2}$.

$\alpha + \beta = \frac{7}{2}, \quad \alpha\beta = \frac{3}{2}$

New sum: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \dfrac{49}{4} - 3 = \dfrac{37}{4}$.

New product: $\alpha^2 \beta^2 = (\alpha\beta)^2 = \dfrac{9}{4}$.

Equation: $x^2 - \dfrac{37}{4}x + \dfrac{9}{4} = 0$I.e. $4x^2 - 37x + 9 = 0$.

Let $t = |x| \geq 0$. The equation becomes $t^2 + 2t - 8 = 0$.

$(t + 4)(t - 2) = 0 \implies t = 2$

(Reject $t = -4$ since $t \geq 0$.)

$|x| = 2 \implies x = 2 \;\mathrm{or}\; x = -2$

For $ax^2 + bx + c \gt 0$ for all real $x$We need $a \gt 0$ and $\Delta \lt 0$.

Condition 1: $k \gt 0$.

Condition 2:

$\Delta = (-2k)^2 - 4(k)(3) = 4k^2 - 12k = 4k(k - 3) \lt 0$

This gives $0 \lt k \lt 3$.

Combined with Condition 1: $0 \lt k \lt 3$.

By the AM-GM inequality (for $x \gt 0$):

$x + \frac{9}{x} \geq 2\sqrt{x \cdot \frac{9}{x}} = 2\sqrt{9} = 6$

Equality holds when $x = \dfrac{9}{x} \implies x^2 = 9 \implies x = 3$ (positive since $x \gt 0$).

Minimum value: $6$Attained at $x = 3$.

Factor: $3x^2 + 2x - 5 = (3x + 5)(x - 1)$.

Roots: $x = -\dfrac{5}{3}$ and $x = 1$. The parabola opens upward ($a = 3 \gt 0$).

The expression is non-positive between the roots:

$-\frac{5}{3} \leq x \leq 1$

$\alpha + \beta = 5, \quad \alpha\beta = 2$

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 125 - 3(2)(5) = 125 - 30 = 95$

Require $p \gt 0$ and $\Delta \lt 0$:

$\Delta = 4p^2 - 4p = 4p(p - 1) \lt 0 \implies 0 \lt p \lt 1$

A perfect square means $\Delta = 0$: $p^2 - 4q = 0 \implies q = \dfrac{p^2}{4}$.

Substituting into $p + q = 9$: $p + \dfrac{p^2}{4} = 9 \implies p^2 + 4p - 36 = 0$.

$p = \frac{-4 \pm \sqrt{16 + 144}}{2} = -2 \pm 2\sqrt{10}$

$q = 9 - p = 9 + 2 \mp 2\sqrt{10} = 11 \mp 2\sqrt{10}$

Answer: $\left(p,\; q\right) = \left(-2 + 2\sqrt{10},\; 11 - 2\sqrt{10}\right)$ or $\left(-2 - 2\sqrt{10},\; 11 + 2\sqrt{10}\right)$.

Let the dimensions be $x$ and $y$ metres. Then $2(x+y) = 120 \implies x + y = 60$So $y = 60 - x$.

$A = xy = x(60 - x) = -x^2 + 60x = -(x - 30)^2 + 900$

Since $-(x-30)^2 \leq 0$ for all real $x$We have $A \leq 900$.

Maximum $A = 900\mathrm{ m}^2$ when $x = 30$Giving $y = 30$.

The field is a $30\mathrm{ m} \times 30\mathrm{ m}$ square.

For real roots: $\Delta = 4a^2 - 16 \geq 0 \implies a^2 \geq 4 \implies a \leq -2$ or $a \geq 2$.

$\alpha + \beta = -2a, \quad \alpha\beta = 4$

$(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 4a^2 - 16$

$\beta - \alpha = 2\sqrt{a^2 - 4}$

(positive since $\beta \gt \alpha$)

$2\sqrt{a^2 - 4} \gt 3 \implies a^2 - 4 \gt \frac{9}{4} \implies a^2 \gt \frac{25}{4} \implies |a| \gt \frac{5}{2}$

Combined with $|a| \geq 2$: $a \lt -\dfrac{5}{2}$ or $a \gt \dfrac{5}{2}$.

Let $y = \dfrac{x^2 - x + 1}{x^2 + x + 1}$. Since $x^2 + x + 1 = \left(x + \dfrac{1}{2}\right)^2 + \dfrac{3}{4} \gt 0$ for all real $x$The denominator never vanishes.

$y(x^2 + x + 1) = x^2 - x + 1 \implies yx^2 + yx + y = x^2 - x + 1$

$(y - 1)x^2 + (y + 1)x + (y - 1) = 0$

For real $x$, $\Delta \geq 0$:

$(y+1)^2 - 4(y-1)^2 \geq 0$

$(y + 1 - 2y + 2)(y + 1 + 2y - 2) \geq 0$

$(-y + 3)(3y - 1) \geq 0 \implies (y - 3)(3y - 1) \leq 0$

$\frac{1}{3} \leq y \leq 3$

Range: $\left[\dfrac{1}{3},\; 3\right]$.

Note that $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = 100 - 2(\alpha\beta)^2$.

$100 - 2(\alpha\beta)^2 = 82 \implies 2(\alpha\beta)^2 = 18 \implies (\alpha\beta)^2 = 9 \implies \alpha\beta = 3 \;\mathrm{or}\; \alpha\beta = -3$

Case $\alpha\beta = 3$: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 10 \implies (\alpha + \beta)^2 - 6 = 10 \implies (\alpha + \beta)^2 = 16 \implies \alpha + \beta = \pm 4$.

So $p = -(\alpha + \beta) = \pm 4$, $q = \alpha\beta = 3$. Two quadratics: $x^2 + 4x + 3 = 0$ and $x^2 - 4x + 3 = 0$.

Case $\alpha\beta = -3$: $(\alpha + \beta)^2 - 2(-3) = 10 \implies (\alpha + \beta)^2 = 4 \implies \alpha + \beta = \pm 2$.

So $p = \mp 2$, $q = -3$. Two quadratics: $x^2 + 2x - 3 = 0$ and $x^2 - 2x - 3 = 0$.

All four pairs are valid solutions.

Setting the two expressions equal:

$x^2 - 3x + 1 = 2x^2 - 5x + 4$

$0 = x^2 - 2x + 3$

$\Delta = 4 - 12 = -8 < 0$

Since the discriminant is negative, the two curves do not intersect.

From the first equation: $y = 5 - x$. Substituting into the second:

$x^2 + (5 - x)^2 = 13$

$x^2 + 25 - 10x + x^2 = 13$

$2x^2 - 10x + 12 = 0$

$x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0$

$x = 2 \implies y = 3$; $x = 3 \implies y = 2$.

Solutions: $(2, 3)$ and $(3, 2)$.

Let $u = \dfrac{x + 1}{x}$. Then $\dfrac{x}{x + 1} = \dfrac{1}{u}$.

$u + \frac{1}{u} = \frac{25}{6}$

$6u^2 + 6 = 25u \implies 6u^2 - 25u + 6 = 0 \implies (2u - 1)(3u - 6) = 0$

$u = \frac{1}{2} \;\text{or}\; u = 2$

Case 1: $\dfrac{x + 1}{x} = \dfrac{1}{2} \implies 2x + 2 = x \implies x = -2$.

Case 2: $\dfrac{x + 1}{x} = 2 \implies x + 1 = 2x \implies x = 1$.

Check: $x = -2$: $\dfrac{-1}{-2} + \dfrac{-2}{-1} = \dfrac{1}{2} + 2 = \dfrac{5}{2} \neq \dfrac{25}{6}$.

Let me redo: $x = -2$: $\dfrac{-2+1}{-2} + \dfrac{-2}{-2+1} = \dfrac{-1}{-2} + \dfrac{-2}{-1} = \dfrac{1}{2} + 2 = \dfrac{5}{2} \neq \dfrac{25}{6}$.

This is incorrect. Let me recheck: $u = 2$ gives $\dfrac{x+1}{x} = 2 \implies x + 1 = 2x \implies x = 1$.

$x = 1$: $\dfrac{2}{1} + \dfrac{1}{2} = \dfrac{5}{2} \neq \dfrac{25}{6}$.

The factorisation $(2u - 1)(3u - 6) = 6u^2 - 12u - 3u + 6 = 6u^2 - 15u + 6 \neq 6u^2 - 25u + 6$.

Correct factorisation: $6u^2 - 25u + 6 = (2u - 3)(3u - 2)$.

$u = \dfrac{3}{2}$ or $u = \dfrac{2}{3}$.

$u = \dfrac{3}{2}$: $\dfrac{x+1}{x} = \dfrac{3}{2} \implies 2x + 2 = 3x \implies x = 2$. Check: $\dfrac{3}{2} + \dfrac{2}{3} = \dfrac{13}{6} \neq \dfrac{25}{6}$.

The original equation $\dfrac{x+1}{x} + \dfrac{x}{x+1} = \dfrac{25}{6}$ with $u + \dfrac{1}{u} = \dfrac{25}{6}$:

$u^2 + 1 = \dfrac{25}{6}u \implies 6u^2 - 25u + 6 = 0$.

$\Delta = 625 - 144 = 481$. $u = \dfrac{25 \pm \sqrt{481}}{12}$.

Since $\sqrt{481} \approx 21.93$:

$u \approx \dfrac{25 + 21.93}{12} \approx 3.91$ or $u \approx \dfrac{25 - 21.93}{12} \approx 0.256$.

This problem has irrational roots. Solutions: $x = \dfrac{1}{u - 1}$.

The range has a maximum of $5$So $a < 0$ (parabola opens downward).

Completing the square: $f(x) = a\!\left(x^2 + \dfrac{4}{a}x\right) + 3 = a\!\left[\left(x + \dfrac{2}{a}\right)^2 - \dfrac{4}{a^2}\right] + 3 = a\!\left(x + \dfrac{2}{a}\right)^2 - \dfrac{4}{a} + 3$.

Maximum value: $3 - \dfrac{4}{a} = 5 \implies -\dfrac{4}{a} = 2 \implies a = -2$.

Verification: $f(x) = -2x^2 + 4x + 3 = -2(x-1)^2 + 5$. Range: $(-\infty, 5]$. Correct.

$\alpha\beta = q = 3$

$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \implies 16 = p^2 - 12 \implies p^2 = 28 \implies p = \pm 2\sqrt{7}$

Answer: $q = 3$, $p = 2\sqrt{7}$ or $p = -2\sqrt{7}$.

$\alpha + \beta = \dfrac{3}{2}$, $\alpha\beta = -2$.

New sum: $\dfrac{1}{\alpha + 1} + \dfrac{1}{\beta + 1} = \dfrac{\alpha + \beta + 2}{\alpha\beta + \alpha + \beta + 1} = \dfrac{3/2 + 2}{-2 + 3/2 + 1} = \dfrac{7/2}{1/2} = 7$.

New product: $\dfrac{1}{(\alpha + 1)(\beta + 1)} = \dfrac{1}{\alpha\beta + \alpha + \beta + 1} = \dfrac{1}{-2 + 3/2 + 1} = \dfrac{1}{1/2} = 2$.

Equation: $x^2 - 7x + 2 = 0$.

For two distinct real roots: $\Delta > 0$:

$\Delta = 4(k-1)^2 - 4(k+5) = 4(k^2 - 2k + 1 - k - 5) = 4(k^2 - 3k - 4) > 0$

$(k - 4)(k + 1) > 0 \implies k < -1 \;\text{or}\; k > 4$

For both roots positive: by Vieta, $\alpha + \beta = -2(k - 1) > 0 \implies k < 1$And $\alpha\beta = k + 5 > 0 \implies k > -5$.

Combining all three conditions: $-5 < k < -1$.

Completing the square: $x^2 - 2x + 3 = (x - 1)^2 + 2$.