Quadratics — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for quadratics.

UT-1: Discriminant with Parameters

Question:

Find the range of values of $k$ for which the equation $(k - 1)x^2 - 2kx + k + 3 = 0$ has two distinct real roots.

Solution:

For two distinct real roots, we need $\Delta > 0$ AND $a \neq 0$.

$a = k - 1$, $b = -2k$, $c = k + 3$.

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

$\Delta > 0 \implies -8k + 12 > 0 \implies k < \dfrac{3}{2}$.

$a \neq 0 \implies k \neq 1$.

Therefore $k \in (-\infty,\; 1) \cup (1,\; \tfrac{3}{2})$.

A common mistake is forgetting the $a \neq 0$ condition (i.e. $k \neq 1$), which would reduce the problem to a linear equation.

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UT-2: Hidden Quadratic in Exponent

Question:

Solve $4^{x+1} - 5 \cdot 2^x + 1 = 0$.

Solution:

Let $u = 2^x$ (so $u > 0$). Then $4^{x+1} = 4 \cdot 4^x = 4u^2$.

$4u^2 - 5u + 1 = 0$

$(4u - 1)(u - 1) = 0$

$u = \frac{1}{4} \quad \text{or} \quad u = 1$

Since $u > 0$, both are valid.

$2^x = \dfrac{1}{4} = 2^{-2} \implies x = -2$.

$2^x = 1 = 2^0 \implies x = 0$.

Therefore $x = -2$ or $x = 0$.

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UT-3: Vieta's Formulas Application

Question:

If $\alpha$ and $\beta$ are the roots of $2x^2 - 6x + 1 = 0$, find the value of $\alpha^2 + \beta^2$ without solving the equation.

Solution:

By Vieta's formulas: $\alpha + \beta = \dfrac{6}{2} = 3$ and $\alpha\beta = \dfrac{1}{2}$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 9 - 2\left(\frac{1}{2}\right) = 9 - 1 = 8$

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UT-4: Completing the Square to Find Range

Question:

Find the range of $f(x) = -3x^2 + 12x - 7$.

Solution:

$f(x) = -3(x^2 - 4x) - 7 = -3(x - 2)^2 + 12 - 7 = -3(x - 2)^2 + 5$

Since $-3(x - 2)^2 \leq 0$ for all $x$, the maximum value is $5$ at $x = 2$.

$\mathrm{ran}(f) = (-\infty,\; 5]$

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UT-5: Factorisation with Non-Integer Coefficients

Question:

Factorise $6x^2 - 7x - 20$ completely.

Solution:

We need $ac = 6 \times (-20) = -120$ and $b = -7$.

Two numbers multiplying to $-120$ and adding to $-7$: $-15$ and $8$.

$6x^2 - 15x + 8x - 20 = 3x(2x - 5) + 4(2x - 5) = (3x + 4)(2x - 5)$

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Integration Tests

Tests synthesis of quadratics with other topics.

IT-1: Quadratics and Inequalities (with Inequalities)

Question:

Find the range of values of $x$ for which $\dfrac{x^2 - 4x + 3}{x^2 - 1} \lt 0$.

Solution:

Factorise: $\dfrac{(x-1)(x-3)}{(x-1)(x+1)} = \dfrac{x - 3}{x + 1}$, provided $x \neq 1$.

$\dfrac{x - 3}{x + 1} \lt 0$.

Critical values: $x = -1$ and $x = 3$. Note $x = 1$ is excluded (makes denominator zero in original).

Sign chart:

Interval	Test	Sign
$x < -1$	$x = -2$	$+$
$-1 < x < 1$	$x = 0$	$-$
$1 < x < 3$	$x = 2$	$-$
$x > 3$	$x = 4$	$+$

The expression is negative for $-1 < x < 1$ or $1 < x < 3$.

Combined: $x \in (-1,\; 1) \cup (1,\; 3)$.

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IT-2: Quadratics and Functions (with Functions)

Question:

Let $f(x) = x^2 - 4x + k$. If the equation $f(x) = 0$ has no real roots and $f(2) > 0$, find the range of $k$.

Solution:

No real roots means $\Delta < 0$:

$\Delta = 16 - 4k < 0 \implies k > 4$

Also $f(2) = 4 - 8 + k = k - 4 > 0 \implies k > 4$.

Both conditions give $k > 4$.

Therefore $k \in (4,\; \infty)$.

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IT-3: Quadratics and Coordinate Geometry (with Coordinate Geometry)

Question:

The line $y = 2x + 1$ intersects the parabola $y = x^2 - 3x + 7$ at points $A$ and $B$. Find the coordinates of $A$ and $B$, and the length of $AB$.

Solution:

Setting equal: $x^2 - 3x + 7 = 2x + 1$

$x^2 - 5x + 6 = 0$

$(x - 2)(x - 3) = 0$

$x = 2 \quad \text{or} \quad x = 3$

When $x = 2$: $y = 2(2) + 1 = 5$. So $A = (2, 5)$.

When $x = 3$: $y = 2(3) + 1 = 7$. So $B = (3, 7)$.

$AB = \sqrt{(3-2)^2 + (7-5)^2} = \sqrt{1 + 4} = \sqrt{5}$

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Worked Examples

WE-1: Sum and Product of Roots Application

Question:

If $\alpha$ and $\beta$ are the roots of $3x^2 - 5x + 2 = 0$, find a quadratic equation whose roots are $\alpha^2$ and $\beta^2$.

Solution:

By Vieta's formulas: $\alpha + \beta = \dfrac{5}{3}$ and $\alpha\beta = \dfrac{2}{3}$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{25}{9} - \frac{4}{3} = \frac{25 - 12}{9} = \frac{13}{9}$

$\alpha^2 \beta^2 = (\alpha\beta)^2 = \frac{4}{9}$

The required equation has sum $= \dfrac{13}{9}$ and product $= \dfrac{4}{9}$:

$x^2 - \frac{13}{9}x + \frac{4}{9} = 0$

$9x^2 - 13x + 4 = 0$

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WE-2: Quadratic Inequality with Parameter

Question:

Find the range of values of $m$ for which the equation $x^2 + 2mx + m^2 - 1 = 0$ has real roots, and find the range of values of $x$ satisfying the inequality $x^2 + 2mx + m^2 - 1 \leq 0$ when $m = 2$.

Solution:

Discriminant: $\Delta = (2m)^2 - 4(m^2 - 1) = 4m^2 - 4m^2 + 4 = 4$.

Since $\Delta = 4 > 0$ for all $m$, the equation always has two distinct real roots.

When $m = 2$: $x^2 + 4x + 3 \leq 0$, i.e. $(x + 1)(x + 3) \leq 0$.

Solution: $-3 \leq x \leq -1$, i.e. $x \in [-3,\; -1]$.

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WE-3: Hidden Quadratic with Substitution

Question:

Solve $x^4 - 5x^2 + 4 = 0$.

Solution:

Let $u = x^2$ ($u \geq 0$).

$u^2 - 5u + 4 = 0$

$(u - 1)(u - 4) = 0$

$u = 1 \quad \text{or} \quad u = 4$

$x^2 = 1 \implies x = 1$ or $x = -1$.

$x^2 = 4 \implies x = 2$ or $x = -2$.

Solution: $x = -2,\; -1,\; 1,\; 2$.

DSE Exam Technique: When solving hidden quadratics, always check the substitution condition ($u \geq 0$) and state all four roots explicitly.

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WE-4: Maximum Value Application

Question:

A farmer has 100 m of fencing to enclose a rectangular field beside a river. No fencing is needed along the river. Find the dimensions that give the maximum area.

Solution:

Let the side parallel to the river have length $x$ and the other two sides have length $y$ each.

$2y + x = 100 \implies y = \dfrac{100 - x}{2} = 50 - \dfrac{x}{2}$.

Area: $A = xy = x\left(50 - \dfrac{x}{2}\right) = 50x - \dfrac{x^2}{2}$.

$A = -\frac{1}{2}(x^2 - 100x) = -\frac{1}{2}(x - 50)^2 + 1250$

Maximum area is $1250$ m$^2$ when $x = 50$ m, $y = 25$ m.

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WE-5: Discriminant Analysis for Tangency

Question:

Find the value of $c$ for which the line $y = 2x + c$ is tangent to the curve $y = x^2 + 3x - 1$.

Solution:

Set equal: $x^2 + 3x - 1 = 2x + c$.

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

For tangency, $\Delta = 0$:

$\Delta = 1 - 4(1)(-(1+c)) = 1 + 4(1+c) = 5 + 4c = 0$

$c = -\frac{5}{4}$

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WE-6: Quadratic with Integer Roots

Question:

Find the integer values of $k$ for which $x^2 + kx + k + 3 = 0$ has integer roots.

Solution:

Let the roots be $\alpha$ and $\beta$ (integers). By Vieta: $\alpha + \beta = -k$ and $\alpha\beta = k + 3$.

Substituting: $\alpha\beta = -(\alpha + \beta) + 3$.

$\alpha\beta + \alpha + \beta = 3$

$(\alpha + 1)(\beta + 1) = 4$

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

$\alpha + 1$	$\beta + 1$	$\alpha$	$\beta$	$k = -(\alpha+\beta)$
1	4	0	3	$-3$
2	2	1	1	$-2$
4	1	3	0	$-3$
$-1$	$-4$	$-2$	$-5$	$7$
$-2$	$-2$	$-3$	$-3$	$6$
$-4$	$-1$	$-5$	$-2$	$7$

Distinct values of $k$: $-3$, $-2$, $6$, $7$.

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WE-7: Using Roots to Form New Equations

Question:

The roots of $x^2 - 7x + 10 = 0$ are $\alpha$ and $\beta$. Find the equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

Solution:

$\alpha + \beta = 7$, $\alpha\beta = 10$.

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7}{10}$

$\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{10}$

Required equation: $x^2 - \dfrac{7}{10}x + \dfrac{1}{10} = 0$, i.e. $10x^2 - 7x + 1 = 0$.

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WE-8: Range of Quadratic Expression

Question:

Find the minimum value of $\dfrac{x^2 + 2x + 5}{x^2 + 2x + 2}$ for real $x$.

Solution:

Let $t = x^2 + 2x + 2 = (x + 1)^2 + 1 \geq 1$.

The expression becomes $\dfrac{t + 3}{t} = 1 + \dfrac{3}{t}$.

Since $t \geq 1$ and $\dfrac{3}{t}$ is decreasing for $t > 0$:

The minimum occurs when $t$ is maximum? No -- $t$ can be arbitrarily large, making $\dfrac{3}{t}$ approach $0$. We need the minimum value, which occurs when $t$ is largest? No. When $t$ increases, $\dfrac{3}{t}$ decreases, so the expression approaches 1 from above.

Actually, since $t \geq 1$ and $\dfrac{3}{t}$ decreases as $t$ increases, the maximum of $\dfrac{3}{t}$ is at $t = 1$:

Minimum of the expression is when $t$ is as large as possible, giving $\dfrac{3}{t} \to 0$.

Wait -- we need the minimum. $1 + \dfrac{3}{t}$ with $t \geq 1$: since $\dfrac{3}{t} \leq 3$ and decreases, the minimum is the infimum $1$ (not attained).

But let us reconsider. The expression $1 + \dfrac{3}{t}$ with $t \geq 1$: since $\dfrac{3}{t}$ ranges from $3$ (at $t = 1$) down to $0$ (as $t \to \infty$), the range is $(1,\; 4]$.

The minimum value approaches $1$ but is never attained.

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Common Pitfalls

1. Forgetting the $a \neq 0$ condition in discriminant problems. When asked about the roots of $ax^2 + bx + c = 0$, if $a$ contains a parameter, always check that $a \neq 0$ separately. If $a = 0$, the equation is linear and has exactly one root.

2. Losing solutions when dividing by an expression containing $x$. When solving $x \cdot f(x) = 0$, you must consider both $x = 0$ and $f(x) = 0$. Dividing by $x$ loses the solution $x = 0$.

3. Incorrect sign when completing the square. A common error is writing $x^2 - 6x = (x - 3)^2 - 9$ but mistakenly writing $-9$ as $+9$. Always verify by expanding back: $(x - 3)^2 = x^2 - 6x + 9$, so $x^2 - 6x = (x - 3)^2 - 9$.

4. Assuming all quadratics can be factorised with integer coefficients. If the discriminant is not a perfect square, the roots are irrational. In such cases, use the quadratic formula and leave answers in surd form (exact values preferred in DSE).

5. Confusing the axis of symmetry with the vertex. For $y = a(x - h)^2 + k$, the axis of symmetry is $x = h$ and the vertex is $(h, k)$. The vertex is a point; the axis of symmetry is a line.

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DSE Exam-Style Questions

DSE-1

(a) Find the range of values of $k$ for which $kx^2 - (k + 3)x + 3 = 0$ has two distinct real roots. (4 marks) (b) For the value of $k$ at the boundary of this range, solve the equation. (2 marks)

Solution:

(a) For two distinct real roots: $\Delta > 0$ and $a \neq 0$.

$a = k$, $b = -(k + 3)$, $c = 3$.

$\Delta = (k + 3)^2 - 4(k)(3) = k^2 + 6k + 9 - 12k = k^2 - 6k + 9 = (k - 3)^2$

$\Delta > 0 \implies (k - 3)^2 > 0 \implies k \neq 3$.

Also $a \neq 0 \implies k \neq 0$.

Therefore $k \in (-\infty,\; 0) \cup (0,\; 3) \cup (3,\; \infty)$.

(b) At the boundary $k = 3$: $\Delta = 0$, one repeated root.

$3x^2 - 6x + 3 = 0 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1$ (repeated root).

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DSE-2

The quadratic equation $x^2 + px + q = 0$ has roots $\alpha$ and $\beta$. It is given that $\alpha^2 + \beta^2 = 10$ and $\alpha^3 + \beta^3 = 28$.

(a) Find the values of $p$ and $q$. (4 marks) (b) Hence solve the equation $x^2 + px + q = 0$. (2 marks)

Solution:

(a) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q = 10$. ... (1)

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = -p^3 + 3pq = 28$. ... (2)

From (1): $p^2 - 2q = 10 \implies q = \dfrac{p^2 - 10}{2}$.

Substituting into (2): $-p^3 + 3p \cdot \dfrac{p^2 - 10}{2} = 28$.

$-p^3 + \frac{3p^3 - 30p}{2} = 28$

$\frac{-2p^3 + 3p^3 - 30p}{2} = 28$

$p^3 - 30p = 56$

$p^3 - 30p - 56 = 0$

By trial: $p = -2$: $-8 + 60 - 56 = -4 \neq 0$.

$p = 7$: $343 - 210 - 56 = 77 \neq 0$.

$p = -4$: $-64 + 120 - 56 = 0$. Yes.

So $p = -4$, $q = \dfrac{16 - 10}{2} = 3$.

(b) $x^2 - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0 \implies x = 1$ or $x = 3$.

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DSE-3

Let $f(x) = x^2 - 2kx + k^2 + k - 3$.

(a) Find the range of values of $k$ for which $f(x) > 0$ for all real $x$. (3 marks) (b) If the minimum value of $f(x)$ is $-1$, find $k$. (3 marks)

Solution:

(a) $f(x) > 0$ for all $x$ requires $\Delta < 0$ (since leading coefficient $= 1 > 0$).

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

$\Delta < 0 \implies -4k + 12 < 0 \implies k > 3$.

(b) Completing the square: $f(x) = (x - k)^2 + k^2 + k - 3 - k^2 = (x - k)^2 + k - 3$.

Minimum value $= k - 3 = -1 \implies k = 2$.

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DSE-4

Solve the inequality $\dfrac{x^2 - 5x + 6}{x^2 - 4} \geq 0$.

Solution:

Factorise: $\dfrac{(x-2)(x-3)}{(x-2)(x+2)} = \dfrac{x - 3}{x + 2}$, provided $x \neq 2$.

Critical values: $x = -2$ (excluded, denominator zero) and $x = 3$ (included, numerator zero). Also $x = 2$ is excluded (denominator zero in original).

Sign chart for $\dfrac{x - 3}{x + 2}$:

Interval	Test	Sign
$x < -2$	$x = -3$	$+$
$-2 < x < 2$	$x = 0$	$-$
$2 < x < 3$	$x = 2.5$	$-$
$x > 3$	$x = 4$	$+$

The expression is non-negative for $x < -2$ or $x \geq 3$, excluding $x = 2$.

Solution: $x \in (-\infty,\; -2) \cup [3,\; \infty)$.

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DSE-5

Given that the equation $2x^2 + 4x + k = 0$ has roots $\alpha$ and $\beta$:

(a) Express $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ in terms of $k$. (2 marks) (b) Find the value of $k$ such that $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = 1$. (4 marks)

Solution:

(a) $\alpha + \beta = -2$, $\alpha\beta = \dfrac{k}{2}$.

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2}{k/2} = -\frac{4}{k}$

(b) $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)^2 - \dfrac{2}{\alpha\beta} = \dfrac{16}{k^2} - \dfrac{4}{k}$.

Setting equal to 1:

$\frac{16}{k^2} - \frac{4}{k} = 1$

$16 - 4k = k^2$

$k^2 + 4k - 16 = 0$

$k = \frac{-4 \pm \sqrt{16 + 64}}{2} = \frac{-4 \pm \sqrt{80}}{2} = -2 \pm 2\sqrt{5}$

Both are valid provided the original equation has real roots: $\Delta = 16 - 8k > 0 \implies k < 2$. Since $-2 + 2\sqrt{5} \approx 2.47 > 2$, only $k = -2 - 2\sqrt{5}$ gives real roots. Check: $\Delta = 16 - 8(-2 - 2\sqrt{5}) = 16 + 16 + 16\sqrt{5} = 32 + 16\sqrt{5} > 0$. Both values give $\Delta > 0$, so both are valid.