Algebra Overview - 21 Questions

Algebra Questions

1. Binomial Theorem

Expand fully: \((3x-2y)^5\).

\[ (3x-2y)^5=\sum_{k=0}^{5}\binom{5}{k}(3x)^{5-k}(-2y)^k =243x^5-810x^4y+1080x^3y^2-720x^2y^3+240xy^4-32y^5. \]

2. Fractions

Write as a single fraction in simplest form: \[ \frac{x-1-\tfrac{6}{x}}{\,x-\tfrac{4}{x}\,}. \]

Multiply top and bottom by \(x\): \[ \frac{x^2-x-6}{x^2-4}=\frac{(x-3)(x+2)}{(x-2)(x+2)}=\frac{x-3}{x-2}, \quad x\neq 0,\pm2. \]

3. Surds

If \(x=\sqrt a+\tfrac1{\sqrt a}\) and \(y=\sqrt a-\tfrac1{\sqrt a}\) with \(a>0\), find \(x^2-y^2\).

\[ x^2-y^2=(x-y)(x+y)=\Big(\tfrac{2}{\sqrt a}\Big)\big(2\sqrt a\big)=4. \]

4. Making and manipulating formulae

Using the constant-acceleration formula \( s = ut + \tfrac{1}{2} a t^2 \) (with \(t>0\)), rearrange to express \(a\) in terms of \(s, u, t\).

\[ s-ut=\tfrac12 a t^2 \;\Rightarrow\; a=\frac{2(s-ut)}{t^2},\qquad t\neq 0. \]

5. Linear simultaneous equations (two variables)

\(\;3x+2y=9,\quad 2x-y=-1.\)

From \(2x-y=-1\Rightarrow y=2x+1\). Substitute: \(3x+2(2x+1)=9\Rightarrow7x=7\Rightarrow x=1\), \(y=3\).

6. Linear simultaneous equations (three variables)

Let \(f,s,d\) be ages of father, son, daughter. Given \(f=3(s+d)\), \(f+s=9d\), \(f+s+d=40\). Find \(f,s,d\).

From \(f+s=9d\) and \(f+s+d=40\Rightarrow10d=40\Rightarrow d=4\). Then \(f+s=36\) and \(f=3(s+4)=3s+12\Rightarrow 4s+12=36\Rightarrow s=6,\; f=30.\)

7. Solving quadratic equations

Solve \(28=x(31+5x)\) by (i) factors, (ii) completing the square, (iii) quadratic formula.

Rearrange \(5x^2+31x-28=0=(5x-4)(x+7)\Rightarrow x=\frac45,-7\). Completing the square and quadratic formula give the same roots.

8. Quadratic graphs

For \(f(x)=2x^2-6x+11\), find the turning point and solve \(f(x)=10\) from your graph.

\[ f(x)=2\big(x^2-3x\big)+11 =2\Big((x-\tfrac32)^2-\tfrac94\Big)+11 =2(x-\tfrac32)^2+ \tfrac{13}{2}. \] Turning point \(\big(\tfrac32,\tfrac{13}{2}\big)\). \(f(x)=10\Rightarrow 2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt7}{2}.\)

9. Nature of quadratic roots

Show roots of \(px^2-(p+q)x+q=0\) are real \(\forall\,p,q\in\mathbb R\) and express them.

\(\Delta=(p+q)^2-4pq=(p-q)^2\ge0\). Roots: \(\displaystyle x=\frac{(p+q)\pm|p-q|}{2p}\) for \(p\neq0\).

10. Linear / non-linear simultaneous equations

\(\;3x-y=1,\quad x^2+4xy=9.\)

\(y=3x-1\Rightarrow x^2+4x(3x-1)=9\Rightarrow 13x^2-4x-9=0\). \(x=\dfrac{4\pm22}{26}\Rightarrow x=1\) or \(-\tfrac{9}{13}\). \(y=2\) or \(-\tfrac{40}{13}\).

11. Rational equations

\(\;\dfrac{1}{x-2}+\dfrac{4}{x+1}=2.\)

\(\dfrac{5x-7}{(x-2)(x+1)}=2\Rightarrow 2x^2-7x+3=0=(2x-1)(x-3)\). \(x=\tfrac12,\,3\) (exclude \(x\neq2,-1\)).

12. Irrational equations

\(\;\sqrt{2x+1}-\sqrt{x-3}=2,\; x\ge3.\)

Square twice: \(x=4\sqrt{x-3}\Rightarrow x^2-16x+48=0\Rightarrow x=4,12\). Both satisfy the original equation.

13. Identities

If \((x+a)^2-(x+b)^2=12x+12\;\forall x\), find \(a,b\).

Difference of squares: \((a-b)(2x+a+b)=12x+12\). \(2(a-b)=12\Rightarrow a-b=6\); \( (a-b)(a+b)=12\Rightarrow a+b=2\). Hence \(a=4,\; b=-2\).

14. Factor Theorem (cubics)

\(f(x)=x^3+ax^2-7x+b\) with factors \(x-1\) and \(x-2\).

\(f(1)=0\Rightarrow a+b-6=0\). \(f(2)=0\Rightarrow 4a-6+b=0\Rightarrow a=0,\;b=6\). \(f(x)=x^3-7x+6=(x-1)(x-2)(x+3)\).

15. Quadratic factor of a cubic

If \(x^2-px+1\) divides \(ax^3+bx+c\;(a\ne0)\), show \(c^2=a(a-b)\).

Write \(ax^3+bx+c=(x^2-px+1)(Ax+B)\). Compare: \(A=a,\;B=ap,\;b=a- ap^2\), \(c=B=ap\). Thus \(c^2=a^2p^2=a\big(a-ap^2\big)=a(a-b).\)

16. Graphing polynomial curves

The graph of \(y=f(x)\) (degree \(4\)) is shown. Given that the curve passes through \((0,-54)\) and has \(x\)-intercepts at \(-3\), \(1\), and \(3\) (with \(x=3\) a double root), find an expression for \(f(x)\).

From the intercepts and multiplicity, write \[ f(x)=k(x+3)(x-1)(x-3)^2. \] Use the point \((0,-54)\): \[ f(0)=-54=k(3)(-1)(9)=-27k \;\Rightarrow\; k=2. \] Hence a correct expression is \[ \boxed{\,f(x)=2(x+3)(x-1)(x-3)^2\,}. \] Expanded form (optional): \[ f(x)=2x^4-8x^3-12x^2+72x-54. \] Check: \(f(0)=-54\) and the roots are \(-3,1,3,3\) as required.

17. Modulus inequalities

Solve \(|2x+5|<3\).

\(-3<2x+5<3\Rightarrow -8<2x<-2\Rightarrow -4<x<-1.\)

18. Abstract inequalities

Prove \(x^2+y^2\ge\frac12(x+y)^2\) for \(x,y\in\mathbb R\).

Since \((x-y)^2\ge0\Rightarrow x^2+y^2\ge2xy\). Hence \(x^2+y^2\ge \tfrac12(x^2+2xy+y^2)=\tfrac12(x+y)^2.\)

19. Rational inequalities

Solve \(\dfrac{5-x}{x-2}<1,\;x\neq2.\)

\(\dfrac{7-2x}{x-2}<0\). Critical points \(x=2,\;\tfrac{7}{2}\). Sign chart gives \(2<x<\tfrac{7}{2}\).

20. Logs and log equations

\(\;\log_5 x = 1 + \log_2\!\big(\frac{3}{2x-1}\big),\;x>\tfrac12.\)

Write \(1=\log_2 2\Rightarrow \log_5 x=\log_2\!\big(\tfrac{6}{2x-1}\big)\). Change of base: \(\dfrac{\ln x}{\ln 5}=\dfrac{\ln\!\big(\tfrac{6}{2x-1}\big)}{\ln 2}\). Solve numerically (no simple closed form): \(x\approx 2.51619\) (3 s.f.: \(2.52\)).

21. Unknown in the index

\(\;2^{2x+1}-17\cdot 2^x+8=0.\)

Let \(u=2^x>0\): \(2u^2-17u+8=0\Rightarrow u=8\) or \(\tfrac12\). Hence \(x=3\) or \(-1\).

HL 2020 – Question 1 (Quadratic & Absolute Value Inequality)

25 marks

▸

(a) Let \( f(x) = x^{2} + 5x + p, \) where \(x \in \mathbb{R}\), \(-3 \le p \le 8\), and \(p \in \mathbb{Z}\).

(i) Find the value of \(p\) for which \(x + 3\) is a factor of \(f(x)\).

(ii) Find the value of \(p\) for which the two roots of \(f(x)=0\) differ by \(3\).

(iii) Find the two values of \(p\) for which the graph of \(y = f(x)\) does not cut the \(x\)-axis.

Solution (a)(i)

If \(x+3\) is a factor of \(f(x)\), then \(x=-3\) is a root and \(f(-3)=0\).

\[ f(-3) = (-3)^2 + 5(-3) + p = 9 - 15 + p = -6 + p. \] So \[ -6 + p = 0 \;\Rightarrow\; p = 6. \]

Solution (a)(ii)

Suppose the roots are \(\alpha\) and \(\alpha + 3\). For \(f(x) = x^2 + 5x + p\), \[ \alpha + (\alpha + 3) = -\frac{5}{1} = -5. \] Hence \[ 2\alpha + 3 = -5 \;\Rightarrow\; 2\alpha = -8 \;\Rightarrow\; \alpha = -4. \] Then the other root is \(\alpha + 3 = -1\).

The product of the roots is \[ \alpha(\alpha+3) = (-4)(-1) = 4, \] and for \(x^2+5x+p\) this product equals \(p\). Therefore \[ p = 4. \]

Solution (a)(iii)

The graph of \(y=f(x)\) does not cut the \(x\)-axis if the quadratic has no real roots, i.e. if the discriminant is negative: \[ \Delta = b^2 - 4ac < 0. \] Here \(a=1\), \(b=5\), \(c=p\), so \[ 5^2 - 4(1)p < 0 \;\Rightarrow\; 25 - 4p < 0 \;\Rightarrow\; 4p > 25 \;\Rightarrow\; p > \tfrac{25}{4}. \]

Given that \(p\) is an integer with \(-3 \le p \le 8\), this gives \[ p \in \{7, 8\}. \]

Marking scheme (a)

(i) Scale 10C (0, 4, 8, 10)

Low partial: shows understanding of \(x+3\) as a factor or \(-3\) as a root, e.g. evaluates \(f(-3)\) or writes \(f(x) = (x+3)(\dots)\).

High partial: forms a correct equation in \(p\) with \(p\) as the only unknown, but with an error in solving.

Full credit: obtains \(p=6\).

(ii) Scale 5C (0, 3, 4, 5)

Low partial: shows understanding that the roots differ by \(3\), e.g. labels roots as \(\alpha\) and \(\alpha\pm3\).

High partial: sets up a relevant equation in \(\alpha\) only (or an equivalent method) but with an error in solving or using the product.

Full credit: obtains \(p = 4\).

(iii) Scale 5C (0, 3, 4, 5)

Low partial: uses or writes the discriminant \(b^2-4ac\) or an equivalent condition for no real roots, or finds one suitable value of \(p\).

High partial: forms a correct inequality in \(p\) alone but with an error in solving or in applying the given range for \(p\).

Full credit: recognises that \(p>25/4\) and hence, using \(-3\le p\le 8\) and \(p\in\mathbb{Z}\), concludes that \(p=7\) or \(p=8\).

(b) Find the range of values of \(x\) for which \( \lvert 2x + 5 \rvert - 1 \le 0, \) where \(x \in \mathbb{R}\).

Solution (b)

Start from \[ \lvert 2x + 5 \rvert - 1 \le 0 \;\Rightarrow\; \lvert 2x + 5 \rvert \le 1. \]

Using the definition of absolute value, \[ \lvert 2x + 5 \rvert \le 1 \;\Longleftrightarrow\; -1 \le 2x + 5 \le 1. \]

Subtract \(5\) from all three parts: \[ -1 - 5 \le 2x \le 1 - 5 \;\Rightarrow\; -6 \le 2x \le -4. \]

Divide through by \(2\): \[ -3 \le x \le -2. \]

Therefore, the required range of values of \(x\) is \[ -3 \le x \le -2. \]

Alternative: From \(\lvert 2x+5\rvert\le1\) we may square both sides, giving \[ (2x+5)^2 \le 1 \;\Rightarrow\; x^2 + 5x + 6 \le 0, \] which factorises as \[ (x+2)(x+3)\le0, \] again leading to \(-3 \le x \le -2\).

Marking scheme (b) — Scale 5D (0, 2, 3, 4, 5)

Low partial (2): writes \((2x+5)^2 \le 1\) or one correct linear inequality such as \(2x+5 \le 1\) or \(-1 \le 2x+5\).

Mid partial (3–4): identifies both linear inequalities \(-1 \le 2x+5 \le 1\) or forms a correct quadratic inequality in \(x\), but does not fully solve for \(x\).

High partial (4): finds the critical values \(-3\) and \(-2\), e.g. by solving \(x^2+5x+6=0\) or by solving \(-6 \le 2x \le -4\), but does not clearly give the final interval.

Full credit (5): gives the complete solution set \(-3 \le x \le -2\). Note: accept the equivalent strict inequality \(-3 < x < -2\) if obtained from a correct method.