Algebra · Paper 1
Quadratics
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Section 1 of 13
Solving by Factorising
A quadratic has an $x^{2}$ term. The cleanest way to solve it — when it factorises — is the CN / Sub method.
CN / Sub — Must learn
For $x^{2} + bx + c = 0$, find two numbers whose CN (product) is $c$ and Sub (sum) is $b$.
Then split the middle term and group.
(i) $x^{2} + 5x + 6 = 0$
CN $= 6$, Sub $= 5$ $\rightarrow$ $2$ and $3$.
$x^{2} + 2x + 3x + 6 = 0$
$x(x + 2) + 3(x + 2) = 0$
$(x + 2)(x + 3) = 0$
$x = -2 \;\;$ or $\;\; x = -3$
(ii) $x^{2} - 7x + 12 = 0$
CN $= 12$, Sub $= -7$ $\rightarrow$ $-3$ and $-4$ (both negative since CN is positive but Sub is negative).
$x^{2} - 3x - 4x + 12 = 0$
$x(x - 3) - 4(x - 3) = 0$
$(x - 3)(x - 4) = 0$
$x = 3 \;\;$ or $\;\; x = 4$
(iii) $2x^{2} - 5x - 3 = 0$
When $a$ ≠ $1$: CN $= a \times c = 2 \times (-3) = -6$, Sub $= -5$ $\rightarrow$ $-6$ and $1$.
$2x^{2} - 6x + x - 3 = 0$
$2x(x - 3) + 1(x - 3) = 0$
$(x - 3)(2x + 1) = 0$
$x = 3 \;\;$ or $\;\; x = -\dfrac{1}{2}$
You try
Solve: $x^{2} + 2x - 15 = 0$
CN $= -15$, Sub $= 2$.
CN $= -15$, Sub $= 2$ $\rightarrow$ $5$ and $-3$
$x^{2} + 5x - 3x - 15 = 0$
$x(x + 5) - 3(x + 5) = 0$
$(x + 5)(x - 3) = 0$
$x = -5 \;\;$ or $\;\; x = 3$
$x = -5 \;\;$ or $\;\; x = 3$
You try
Solve: $3x^{2} + 11x + 6 = 0$
CN $= 3 \times 6 = 18$, Sub $= 11$.
CN $= 18$, Sub $= 11$ $\rightarrow$ $9$ and $2$
$3x^{2} + 9x + 2x + 6 = 0$
$3x(x + 3) + 2(x + 3) = 0$
$(x + 3)(3x + 2) = 0$
$x = -3 \;\;$ or $\;\; x = -\dfrac{2}{3}$
$x = -3 \;\;$ or $\;\; x = -\dfrac{2}{3}$
You try
Solve: $x^{2} - 9 = 0$ (difference of two squares — no middle term)
$a^{2} - b^{2} = (a - b)(a + b)$.
$(x - 3)(x + 3) = 0$
$x = 3 \;\;$ or $\;\; x = -3$
$x = \pm 3$
Section 2 of 13
The Quadratic Formula
When a quadratic doesn't factorise nicely, use the formula. It works on every quadratic.
The formula — Must learn
For $ax^{2} + bx + c = 0$: $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
It's on page 20 of the log tables. The $\pm$ gives the two answers.
(i) $2x^{2} - 7x + 3 = 0$ — check with formula
$a = 2$, $b = -7$, $c = 3$
$x = \dfrac{-(-7) \pm \sqrt{(-7)^{2} - 4(2)(3)}}{2(2)}$
$x = \dfrac{7 \pm \sqrt{49 - 24}}{4} = \dfrac{7 \pm \sqrt{25}}{4} = \dfrac{7 \pm 5}{4}$
$x = \dfrac{12}{4} = 3 \;\;$ or $\;\; x = \dfrac{2}{4} = \dfrac{1}{2}$
It factorises: $(x - 3)(2x - 1) = 0$. Same answers — the formula always works.
(ii) $x^{2} - 5x + 2 = 0$ — give $x$ to 2 d.p.
$a = 1$, $b = -5$, $c = 2$. CN $= 2$, Sub $= -5$ — no whole numbers work, so use the formula.
$x = \dfrac{5 \pm \sqrt{25 - 8}}{2} = \dfrac{5 \pm \sqrt{17}}{2}$
$x = 4.56 \;\;$ or $\;\; x = 0.44$ (2 d.p.)
You try
Solve: $x^{2} + 4x - 6 = 0$ — give $x$ to 2 d.p.
$a = 1$, $b = 4$, $c = -6$. Won't factorise.
$x = \dfrac{-4 \pm \sqrt{16 - 4(1)(-6)}}{2}$
$x = \dfrac{-4 \pm \sqrt{40}}{2}$
$x = 1.16 \;\;$ or $\;\; x = -5.16$
$x = 1.16 \;\;$ or $\;\; x = -5.16$
You try
Solve: $3x^{2} - 8x + 2 = 0$ — give $x$ to 2 d.p.
$a = 3$, $b = -8$, $c = 2$.
$x = \dfrac{8 \pm \sqrt{64 - 24}}{6} = \dfrac{8 \pm \sqrt{40}}{6}$
$x = 2.39 \;\;$ or $\;\; x = 0.28$
$x = 2.39 \;\;$ or $\;\; x = 0.28$
Section 3 of 13
Surd-Form Answers
Sometimes you're asked for the answer in surd form — leave the $\sqrt{\;}$ in the answer rather than rounding.
Tidy the surd: $\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}$, $\sqrt{12} = 2\sqrt{3}$, $\sqrt{50} = 5\sqrt{2}$ — see your Surds lesson.
(i) $x^{2} - 4x + 1 = 0$ — in surd form
$x = \dfrac{4 \pm \sqrt{16 - 4}}{2} = \dfrac{4 \pm \sqrt{12}}{2}$
Tidy: $\sqrt{12} = 2\sqrt{3}$.
$x = \dfrac{4 \pm 2\sqrt{3}}{2}$
Divide every term on top by the bottom:
$x = 2 \pm \sqrt{3}$
(ii) $2x^{2} + 2x - 3 = 0$ — in surd form
$x = \dfrac{-2 \pm \sqrt{4 + 24}}{4} = \dfrac{-2 \pm \sqrt{28}}{4}$
$\sqrt{28} = 2\sqrt{7}$.
$x = \dfrac{-2 \pm 2\sqrt{7}}{4}$
Take out $2$ on top, then cancel:
$x = \dfrac{2(-1 \pm \sqrt{7})}{4}$
$x = \dfrac{-1 \pm \sqrt{7}}{2}$
Don't cancel through a $+$ sign
$\dfrac{4 + 2\sqrt{3}}{2}$ ≠ $2 + 2\sqrt{3}$
You must divide every term on top by the bottom — or take out a common factor and cancel.
You try
Solve in surd form: $x^{2} - 6x + 2 = 0$
$\sqrt{36 - 8} = \sqrt{28} = 2\sqrt{7}$.
$x = \dfrac{6 \pm \sqrt{28}}{2} = \dfrac{6 \pm 2\sqrt{7}}{2}$
$x = \dfrac{2(3 \pm \sqrt{7})}{2}$
$x = 3 \pm \sqrt{7}$
$x = 3 \pm \sqrt{7}$
You try
Solve in surd form: $3x^{2} - 4x - 1 = 0$
$\sqrt{16 + 12} = \sqrt{28} = 2\sqrt{7}$.
$x = \dfrac{4 \pm \sqrt{16 + 12}}{6} = \dfrac{4 \pm \sqrt{28}}{6}$
$x = \dfrac{4 \pm 2\sqrt{7}}{6} = \dfrac{2(2 \pm \sqrt{7})}{6}$
$x = \dfrac{2 \pm \sqrt{7}}{3}$
$x = \dfrac{2 \pm \sqrt{7}}{3}$
Section 4 of 13
Completing the Square
Completing the square rewrites $x^{2} + bx + c$ as $(x + p)^{2} + k$. It's the engine behind min/max points, sketching, and another way to solve quadratics.
The trick — Must learn
Take half the coefficient of $x$, square it, add and subtract it.
$x^{2} + bx \;=\; \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2}$
(i) Complete the square: $x^{2} + 6x$
Half of $6$ is $3$. Square it: $9$.
$x^{2} + 6x = (x + 3)^{2} - 9$
Check: $(x + 3)^{2} - 9 = x^{2} + 6x + 9 - 9 = x^{2} + 6x$. ✓
(ii) Complete the square: $x^{2} - 8x$
Half of $-8$ is $-4$. Square it: $16$.
$x^{2} - 8x = (x - 4)^{2} - 16$
(iii) Complete the square: $x^{2} + 6x + 11$
Keep the constant. Half of $6$ is $3$, squared is $9$.
$x^{2} + 6x + 11 = (x + 3)^{2} - 9 + 11$
$= (x + 3)^{2} + 2$
(iv) Complete the square: $x^{2} - 10x + 4$
Half of $-10$ is $-5$, squared is $25$.
$x^{2} - 10x + 4 = (x - 5)^{2} - 25 + 4$
$= (x - 5)^{2} - 21$
(v) When the half is a fraction: $x^{2} + 5x$
Half of $5$ is $\dfrac{5}{2}$. Squared: $\dfrac{25}{4}$.
$x^{2} + 5x = \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4}$
You try
Complete the square: $x^{2} - 4x + 1$
Half of $-4$ is $-2$, squared is $4$.
$x^{2} - 4x + 1 = (x - 2)^{2} - 4 + 1$
$= (x - 2)^{2} - 3$
$(x - 2)^{2} - 3$
You try
Complete the square: $x^{2} + 8x - 5$
Half of $8$ is $4$, squared is $16$.
$x^{2} + 8x - 5 = (x + 4)^{2} - 16 - 5$
$= (x + 4)^{2} - 21$
$(x + 4)^{2} - 21$
You try
Complete the square: $x^{2} - 3x + 1$
Half of $-3$ is $-\dfrac{3}{2}$, squared is $\dfrac{9}{4}$.
$x^{2} - 3x + 1 = \left(x - \dfrac{3}{2}\right)^{2} - \dfrac{9}{4} + 1$
$= \left(x - \dfrac{3}{2}\right)^{2} - \dfrac{5}{4}$
$\left(x - \dfrac{3}{2}\right)^{2} - \dfrac{5}{4}$
Section 5 of 13
Solving by Completing the Square
Once you've written it as $(x + p)^{2} + k = 0$, isolate the squared bracket, then take square roots — $\pm$ on the right.
(i) Solve in surd form: $x^{2} - 6x + 2 = 0$
Complete the square first.
$(x - 3)^{2} - 9 + 2 = 0$
$(x - 3)^{2} = 7$
Square root both sides — remember the $\pm$:
$x - 3 = \pm\sqrt{7}$
$x = 3 \pm \sqrt{7}$
Same answer as the formula gave us in Section 3 — different road, same destination.
(ii) Solve in surd form: $x^{2} + 4x - 1 = 0$
$(x + 2)^{2} - 4 - 1 = 0$
$(x + 2)^{2} = 5$
$x + 2 = \pm\sqrt{5}$
$x = -2 \pm \sqrt{5}$
You try
Solve by completing the square (surd form): $x^{2} - 2x - 4 = 0$
Half of $-2$ is $-1$, squared is $1$.
$(x - 1)^{2} - 1 - 4 = 0$
$(x - 1)^{2} = 5$
$x - 1 = \pm\sqrt{5}$
$x = 1 \pm \sqrt{5}$
$x = 1 \pm \sqrt{5}$
You try
Solve by completing the square: $x^{2} + 6x + 7 = 0$
Half of $6$ is $3$, squared is $9$.
$(x + 3)^{2} - 9 + 7 = 0$
$(x + 3)^{2} = 2$
$x + 3 = \pm\sqrt{2}$
$x = -3 \pm \sqrt{2}$
$x = -3 \pm \sqrt{2}$
Section 6 of 13
Quadratics with Fractions
When $x$ is on the bottom, you usually get a quadratic after clearing the fractions. Multiply every term by the LCD, then solve.
Don't forget — check the bottom
Any answer that makes a denominator $= 0$ must be thrown out at the end.
(i) Solve: $\dfrac{6}{x} + \dfrac{6}{x + 5} = 1$
LCD is $x(x + 5)$. Multiply every term:
$6(x + 5) + 6x = x(x + 5)$
$6x + 30 + 6x = x^{2} + 5x$
$12x + 30 = x^{2} + 5x$
$x^{2} - 7x - 30 = 0$
CN $= -30$, Sub $= -7$ $\rightarrow$ $-10$ and $3$.
$(x - 10)(x + 3) = 0$
$x = 10 \;\;$ or $\;\; x = -3$
Check: neither makes $x = 0$ nor $x + 5 = 0$. Both valid.
(ii) Solve: $x + \dfrac{6}{x} = 5$
Multiply every term by $x$:
$x^{2} + 6 = 5x$
$x^{2} - 5x + 6 = 0$
$(x - 2)(x - 3) = 0$
$x = 2 \;\;$ or $\;\; x = 3$
(iii) A trap: $\dfrac{3}{x - 2} + \dfrac{2}{x} = 2$
LCD is $x(x - 2)$. Multiply every term:
$3x + 2(x - 2) = 2x(x - 2)$
$3x + 2x - 4 = 2x^{2} - 4x$
$2x^{2} - 9x + 4 = 0$
CN $= 2 \cdot 4 = 8$, Sub $= -9$ $\rightarrow$ $-8$ and $-1$.
$2x^{2} - 8x - x + 4 = 0$
$2x(x - 4) - 1(x - 4) = 0$
$(x - 4)(2x - 1) = 0$
$x = 4 \;\;$ or $\;\; x = \dfrac{1}{2}$
Check: neither makes $x - 2 = 0$ or $x = 0$. Both valid.
You try
Solve: $\dfrac{1}{x} + \dfrac{1}{x + 1} = \dfrac{5}{6}$
LCD is $6x(x + 1)$.
$6(x + 1) + 6x = 5x(x + 1)$
$6x + 6 + 6x = 5x^{2} + 5x$
$5x^{2} - 7x - 6 = 0$
CN $= -30$, Sub $= -7$ $\rightarrow$ $-10, 3$
$5x^{2} - 10x + 3x - 6 = 0$
$5x(x - 2) + 3(x - 2) = 0$
$(x - 2)(5x + 3) = 0$
$x = 2 \;\;$ or $\;\; x = -\dfrac{3}{5}$
$x = 2 \;\;$ or $\;\; x = -\dfrac{3}{5}$
You try
Solve: $x - \dfrac{12}{x} = 1$
Multiply every term by $x$.
$x^{2} - 12 = x$
$x^{2} - x - 12 = 0$
$(x - 4)(x + 3) = 0$
$x = 4 \;\;$ or $\;\; x = -3$
$x = 4 \;\;$ or $\;\; x = -3$
Section 7 of 13
Quadratics by Substitution
Some equations look bigger than quadratics but become quadratic if you substitute. The clue: one power is exactly double the other.
When to substitute
If you see $x^{4}$ and $x^{2}$ → let $y = x^{2}$.
If you see $x$ and $\sqrt{x}$ → let $y = \sqrt{x}$.
If you see $3^{2x}$ and $3^{x}$ → let $y = 3^{x}$.
Always finish by going back to $x$.
(i) $x^{4} - 13x^{2} + 36 = 0$
Let $y = x^{2}$. Then $x^{4} = y^{2}$.
$y^{2} - 13y + 36 = 0$
$(y - 4)(y - 9) = 0$
$y = 4 \;\;$ or $\;\; y = 9$
Now back to $x$ — and remember $\pm$ when square-rooting:
$x^{2} = 4 \;\;\Rightarrow\;\; x = \pm 2$
$x^{2} = 9 \;\;\Rightarrow\;\; x = \pm 3$
$x = \pm 2, \;\; \pm 3$ (four solutions)
(ii) $x - 5\sqrt{x} + 6 = 0$
Let $y = \sqrt{x}$. Then $x = y^{2}$.
$y^{2} - 5y + 6 = 0$
$(y - 2)(y - 3) = 0$
$y = 2 \;\;$ or $\;\; y = 3$
Back to $x$:
$\sqrt{x} = 2 \;\;\Rightarrow\;\; x = 4$
$\sqrt{x} = 3 \;\;\Rightarrow\;\; x = 9$
$x = 4 \;\;$ or $\;\; x = 9$
No $\pm$ here — $\sqrt{x}$ is only the positive root.
(iii) $3^{2x} - 10 \cdot 3^{x} + 9 = 0$
Notice $3^{2x} = (3^{x})^{2}$. Let $y = 3^{x}$.
$y^{2} - 10y + 9 = 0$
$(y - 1)(y - 9) = 0$
$y = 1 \;\;$ or $\;\; y = 9$
Back to $x$:
$3^{x} = 1 \;\;\Rightarrow\;\; x = 0$ (anything to the $0$ is $1$)
$3^{x} = 9 \;\;\Rightarrow\;\; x = 2$
$x = 0 \;\;$ or $\;\; x = 2$
You try
Solve: $x^{4} - 10x^{2} + 9 = 0$
Let $y = x^{2}$.
$y^{2} - 10y + 9 = 0$
$(y - 1)(y - 9) = 0$
$y = 1 \;\Rightarrow\; x^{2} = 1 \;\Rightarrow\; x = \pm 1$
$y = 9 \;\Rightarrow\; x^{2} = 9 \;\Rightarrow\; x = \pm 3$
$x = \pm 1, \;\; \pm 3$
$x = \pm 1, \;\; \pm 3$
You try
Solve: $2^{2x} - 6 \cdot 2^{x} + 8 = 0$
Let $y = 2^{x}$.
$y^{2} - 6y + 8 = 0$
$(y - 2)(y - 4) = 0$
$2^{x} = 2 \;\Rightarrow\; x = 1$
$2^{x} = 4 \;\Rightarrow\; x = 2$
$x = 1 \;\;$ or $\;\; x = 2$
$x = 1 \;\;$ or $\;\; x = 2$
Section 8 of 13
Nature of Roots — The Discriminant
Inside the quadratic formula sits a number that tells us, before solving, what kind of roots the equation has. We call it the discriminant.
The discriminant — Must learn
$\Delta = b^{2} - 4ac$
$\Delta > 0$ → two real, distinct roots (cuts $x$-axis twice)
$\Delta = 0$ → one real, repeated root (touches $x$-axis once)
$\Delta < 0$ → no real roots (misses $x$-axis)
(i) Describe the roots of $\;x^{2} - 5x + 3 = 0$
$a = 1$, $b = -5$, $c = 3$.
$\Delta = (-5)^{2} - 4(1)(3) = 25 - 12 = 13$
$\Delta > 0$ → two real, distinct roots.
(ii) Describe the roots of $\;x^{2} - 6x + 9 = 0$
$\Delta = (-6)^{2} - 4(1)(9) = 36 - 36 = 0$
$\Delta = 0$ → one real, repeated root.
Check: $x^{2} - 6x + 9 = (x - 3)^{2} = 0$, so $x = 3$ (just once). ✓
(iii) Describe the roots of $\;2x^{2} + x + 1 = 0$
$\Delta = (1)^{2} - 4(2)(1) = 1 - 8 = -7$
$\Delta < 0$ → no real roots.
You try
Describe the roots of: $3x^{2} - 4x + 2 = 0$
$\Delta = b^{2} - 4ac$.
$\Delta = (-4)^{2} - 4(3)(2) = 16 - 24 = -8$
$\Delta < 0$ → no real roots.
No real roots ($\Delta = -8$).
You try
Describe the roots of: $4x^{2} - 12x + 9 = 0$
Compute $\Delta = b^{2} - 4ac$.
$\Delta = (-12)^{2} - 4(4)(9) = 144 - 144 = 0$
$\Delta = 0$ → one repeated root.
One repeated root ($\Delta = 0$).
Section 9 of 13
Finding $k$ from the Roots
A quadratic with a hidden parameter $k$ (or $m$, $p$, ...) is fixed by stating what kind of roots it has. Use the discriminant.
(i) For what value of $k$ does $\;x^{2} + kx + 9 = 0$ have equal roots?
"Equal roots" means $\Delta = 0$.
$a = 1$, $b = k$, $c = 9$
$\Delta = k^{2} - 4(1)(9) = 0$
$k^{2} = 36$
$k = \pm 6$
(ii) For what value of $k$ does $\;kx^{2} - 6x + 1 = 0$ have one repeated root?
$a = k$, $b = -6$, $c = 1$
$\Delta = (-6)^{2} - 4(k)(1) = 0$
$36 - 4k = 0$
$k = 9$
(iii) Find $k$ if $\;x^{2} - 4x + k = 0$ has real roots.
"Real roots" means $\Delta \geq 0$.
$(-4)^{2} - 4(1)(k) \geq 0$
$16 - 4k \geq 0$
$-4k \geq -16$
Divide by $-4$ — flip the inequality (signs change ⇒ signs change):
$k \leq 4$
(iv) Find $k$ if $\;x^{2} + (k - 1)x + 4 = 0$ has equal roots.
$\Delta = (k - 1)^{2} - 4(1)(4) = 0$
$(k - 1)^{2} = 16$
$k - 1 = \pm 4$
$k = 5 \;\;$ or $\;\; k = -3$
You try
Find $k$ if $x^{2} + 8x + k = 0$ has one repeated root.
Set $\Delta = 0$.
$\Delta = 64 - 4k = 0$
$4k = 64$
$k = 16$
$k = 16$
You try
Find $k$ if $2x^{2} + kx + 8 = 0$ has equal roots.
$a = 2$, $b = k$, $c = 8$. Set $\Delta = 0$.
$\Delta = k^{2} - 4(2)(8) = 0$
$k^{2} = 64$
$k = \pm 8$
$k = \pm 8$
You try
Find the values of $k$ for which $x^{2} - 2kx + 9 = 0$ has no real roots.
"No real roots" means $\Delta < 0$. Here $b = -2k$.
$\Delta = (-2k)^{2} - 4(1)(9) < 0$
$4k^{2} - 36 < 0$
$k^{2} < 9$
$-3 < k < 3$
$-3 < k < 3$
Section 10 of 13
Forming a Quadratic from Its Roots
Given two roots, build the quadratic that has them. There are two ways — both worth knowing.
Sum & Product method — Must learn
If the roots are $\alpha$ and $\beta$, the quadratic is:
$x^{2} - (\alpha + \beta)x + \alpha\beta = 0$
Read: $x^{2} - (\text{sum})x + (\text{product}) = 0$.
(i) Form a quadratic with roots $\;3$ and $\;-5$.
Way 1 — factor form:
$(x - 3)(x + 5) = 0$
$x^{2} + 5x - 3x - 15 = 0$
$x^{2} + 2x - 15 = 0$
Way 2 — sum & product:
Sum $= 3 + (-5) = -2$
Product $= (3)(-5) = -15$
$x^{2} - (-2)x + (-15) = 0$
$x^{2} + 2x - 15 = 0$ ✓
(ii) Form a quadratic with roots $\;\dfrac{1}{2}$ and $\;-3$.
Sum $= \dfrac{1}{2} + (-3) = -\dfrac{5}{2}$
Product $= \dfrac{1}{2} \cdot (-3) = -\dfrac{3}{2}$
$x^{2} - \left(-\dfrac{5}{2}\right)x + \left(-\dfrac{3}{2}\right) = 0$
$x^{2} + \dfrac{5}{2}x - \dfrac{3}{2} = 0$
Multiply every term by $2$ to clear the fractions:
$2x^{2} + 5x - 3 = 0$
(iii) Form a quadratic with roots $\;2 + \sqrt{3}\;$ and $\;2 - \sqrt{3}$.
Surd roots come in conjugate pairs — sum and product clean up.
Sum $= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$
Product $= (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$
$x^{2} - 4x + 1 = 0$
You try
Form a quadratic with roots $-2$ and $7$.
Sum and product.
Sum $= -2 + 7 = 5$
Product $= (-2)(7) = -14$
$x^{2} - 5x - 14 = 0$
$x^{2} - 5x - 14 = 0$
You try
Form a quadratic with roots $\dfrac{2}{3}$ and $-1$. Give whole-number coefficients.
Sum $= -\dfrac{1}{3}$, Product $= -\dfrac{2}{3}$.
Sum $= \dfrac{2}{3} - 1 = -\dfrac{1}{3}$
Product $= \dfrac{2}{3} \cdot (-1) = -\dfrac{2}{3}$
$x^{2} + \dfrac{1}{3}x - \dfrac{2}{3} = 0$
Multiply by $3$:
$3x^{2} + x - 2 = 0$
$3x^{2} + x - 2 = 0$
You try
Form a quadratic with roots $3 + \sqrt{2}$ and $3 - \sqrt{2}$.
Conjugate pair — sum kills the surd, product is a difference of squares.
Sum $= 6$
Product $= 9 - 2 = 7$
$x^{2} - 6x + 7 = 0$
$x^{2} - 6x + 7 = 0$
Section 11 of 13
Min/Max Point by Completing the Square
Every quadratic has either a lowest point ($a > 0$, U-shape) or a highest point ($a < 0$, ∩-shape). Completing the square reveals it instantly.
The turning point — Must learn
In the form $\;y = (x + p)^{2} + k\;$ the turning point is at:
$(\,-p, \;\; k\,)$
Because $(x + p)^{2} \geq 0$, the smallest value $y$ can take is $k$, and that happens when $x + p = 0$.
(i) Find the min point of $\;y = x^{2} - 6x + 11$
Complete the square: half of $-6$ is $-3$, squared is $9$.
$y = (x - 3)^{2} - 9 + 11$
$y = (x - 3)^{2} + 2$
The bracket is smallest ($= 0$) when $x = 3$, and then $y = 2$.
Min point: $(3, \; 2)$
(ii) Find the min point of $\;y = x^{2} + 4x - 1$
$y = (x + 2)^{2} - 4 - 1 = (x + 2)^{2} - 5$
Min point: $(-2, \; -5)$
(iii) A max-point example: $\;y = -x^{2} + 4x + 1$
Negative $a$ means a "frown" — there's a max. Factor out $-1$ from the $x$ terms:
$y = -(x^{2} - 4x) + 1$
$y = -[(x - 2)^{2} - 4] + 1$
$y = -(x - 2)^{2} + 4 + 1$
$y = -(x - 2)^{2} + 5$
$-(x - 2)^{2}$ is biggest ($= 0$) when $x = 2$, and then $y = 5$.
Max point: $(2, \; 5)$
You try
Find the min point of $y = x^{2} - 8x + 10$.
Half of $-8$ is $-4$, squared is $16$.
$y = (x - 4)^{2} - 16 + 10$
$y = (x - 4)^{2} - 6$
Min point: $(4, \; -6)$
$(4, \; -6)$
You try
Find the min point of $y = x^{2} + 2x + 5$.
Half of $2$ is $1$, squared is $1$.
$y = (x + 1)^{2} - 1 + 5 = (x + 1)^{2} + 4$
Min point: $(-1, \; 4)$
$(-1, \; 4)$
Section 12 of 13
When $a$ ≠ $1$
When the coefficient of $x^{2}$ isn't $1$, factor it out of the $x$ terms first, complete the square inside the bracket, then tidy up.
(i) Find the min point of $\;y = 2x^{2} - 8x + 3$
Factor $2$ out of the $x^{2}$ and $x$ terms only:
$y = 2(x^{2} - 4x) + 3$
Inside the bracket: half of $-4$ is $-2$, squared is $4$.
$y = 2[(x - 2)^{2} - 4] + 3$
Multiply the $2$ back in:
$y = 2(x - 2)^{2} - 8 + 3$
$y = 2(x - 2)^{2} - 5$
Min point: $(2, \; -5)$
(ii) Express $\;3x^{2} + 12x + 7\;$ in completed-square form.
$3x^{2} + 12x + 7 = 3(x^{2} + 4x) + 7$
$= 3[(x + 2)^{2} - 4] + 7$
$= 3(x + 2)^{2} - 12 + 7$
$= 3(x + 2)^{2} - 5$
(iii) Find the max point of $\;y = -2x^{2} + 8x + 1$
Factor $-2$ out of the $x^{2}$ and $x$ terms:
$y = -2(x^{2} - 4x) + 1$
$y = -2[(x - 2)^{2} - 4] + 1$
$y = -2(x - 2)^{2} + 8 + 1$
$y = -2(x - 2)^{2} + 9$
Max point: $(2, \; 9)$
Watch the sign
When you take $-2$ out of $\;-2x^{2} + 8x$, the inside becomes $x^{2} - 4x$ — not $x^{2} + 4x$. Sign flips when factoring out a negative.
You try
Find the min point of $y = 2x^{2} + 8x + 1$.
Factor $2$ from $x$ terms first.
$y = 2(x^{2} + 4x) + 1$
$y = 2[(x + 2)^{2} - 4] + 1$
$y = 2(x + 2)^{2} - 8 + 1$
$y = 2(x + 2)^{2} - 7$
Min point: $(-2, \; -7)$
$(-2, \; -7)$
You try
Find the max point of $y = -x^{2} + 6x - 4$.
Factor $-1$ from $x$ terms.
$y = -(x^{2} - 6x) - 4$
$y = -[(x - 3)^{2} - 9] - 4$
$y = -(x - 3)^{2} + 9 - 4$
$y = -(x - 3)^{2} + 5$
Max point: $(3, \; 5)$
$(3, \; 5)$
Section 13 of 13
Sketching Quadratics
A quick, clean sketch needs four pieces of information. Get all four, then draw.
The four-point method — Must learn
1. Shape — $a > 0$ is U, $a < 0$ is ∩.
2. Roots — where the curve cuts the $x$-axis ($y = 0$).
3. $y$-intercept — the $c$ value ($x = 0$).
4. Turning point — from completing the square (or by symmetry: $x$-coordinate is the midpoint of the roots).
(i) Sketch $\;y = x^{2} - 2x - 8$
1. Shape: $a = 1 > 0$ → U-shape.
2. Roots (set $y = 0$):
$x^{2} - 2x - 8 = 0$
$(x - 4)(x + 2) = 0$
Cuts $x$-axis at $(-2, 0)$ and $(4, 0)$.
3. $y$-intercept (set $x = 0$):
$y = -8$ → cuts $y$-axis at $(0, -8)$.
4. Turning point — midpoint of roots is $\dfrac{-2 + 4}{2} = 1$. Sub in:
$y = (1)^{2} - 2(1) - 8 = -9$
Min point: $(1, \; -9)$.
All four pieces — draw a U through $(-2, 0)$ and $(4, 0)$, $y$-intercept $-8$, turning point $(1, -9)$.
(ii) Sketch $\;y = -x^{2} + 4x - 3$
1. Shape: $a = -1 < 0$ → ∩-shape.
2. Roots:
$-x^{2} + 4x - 3 = 0 \;\;\Rightarrow\;\; x^{2} - 4x + 3 = 0$
$(x - 1)(x - 3) = 0$ → cuts $x$-axis at $(1, 0)$ and $(3, 0)$.
3. $y$-intercept: $y = -3$ → $(0, -3)$.
4. Turning point: midpoint of roots $= 2$. $y = -(2)^{2} + 4(2) - 3 = 1$.
Max point: $(2, \; 1)$.
(iii) What if there are no real roots? $\;y = x^{2} + 2x + 5$
$\Delta = 4 - 20 = -16 < 0$ → no $x$-axis crossings.
The curve sits entirely above the $x$-axis. Use the turning point and $y$-intercept to sketch.
Turning point: $(x + 1)^{2} + 4$ → $(-1, 4)$.
$y$-intercept: $(0, 5)$.
U-shape, min $(-1, 4)$, $y$-intercept $5$, never touches the $x$-axis.
You try
For $y = x^{2} - 4x - 5$, find the roots, $y$-intercept, and turning point.
Factorise first.
Roots: $(x - 5)(x + 1) = 0 \;\Rightarrow\; x = 5, \; -1$
$y$-intercept: $y = -5 \;\Rightarrow\; (0, -5)$
Midpoint of roots: $2$. $y = 4 - 8 - 5 = -9$.
Roots $(5, 0), (-1, 0)$; $y$-int $(0, -5)$; min $(2, -9)$
Roots $-1, 5$; $y$-int $-5$; min $(2, -9)$
You try
For $y = -x^{2} - 2x + 8$, find the roots, $y$-intercept, and turning point.
Shape is ∩ (max). Multiply both sides by $-1$ to factorise roots.
Roots: $x^{2} + 2x - 8 = 0 \;\Rightarrow\; (x + 4)(x - 2) = 0$
$x = -4, \; 2$
$y$-intercept: $(0, 8)$
Midpoint: $-1$. $y = -1 + 2 + 8 = 9$.
Roots $-4, 2$; $y$-int $8$; max $(-1, 9)$
Roots $-4, 2$; $y$-int $8$; max $(-1, 9)$
That's Quadratics.
You can now factorise, use the formula in decimal or surd form, complete the square, handle fractions and substitutions, work with the discriminant, build quadratics from their roots, find turning points, and sketch.