1. Solve $x^2-7x+12=0$ by factorization. - The problem is to find $x$ such that $x^2-7x+12=0$. - Factorization formula: $ax^2+bx+c=0$ can be factored as $(x-p)(x-q)=0$ where $p$ and $q$ satisfy $p+q=-\frac{b}{a}$ and $pq=\frac{c}{a}$. - Here, $a=1$, $b=-7$, $c=12$. - Find two numbers that multiply to 12 and add to -7: -3 and -4. - So, $x^2-7x+12=(x-3)(x-4)=0$. - Set each factor to zero: $x-3=0$ or $x-4=0$. - Solutions: $x=3$ or $x=4$. 2. Solve $2x^2+3x-5=0$ using the quadratic formula. - Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. - Here, $a=2$, $b=3$, $c=-5$. - Calculate discriminant: $\Delta=b^2-4ac=3^2-4\times2\times(-5)=9+40=49$. - Square root of discriminant: $\sqrt{49}=7$. - Solutions: $x=\frac{-3\pm7}{2\times2} = \frac{-3\pm7}{4}$. - Two roots: $x=\frac{-3+7}{4}=1$ and $x=\frac{-3-7}{4}=-2.5$. 3. Find the roots of $x^2+4x+4=0$ and state their nature. - Discriminant: $\Delta=4^2-4\times1\times4=16-16=0$. - Since $\Delta=0$, roots are real and equal. - Roots: $x=\frac{-4}{2}= -2$. - Both roots are $-2$. 4. If the roots of $ax^2+bx+c=0$ are equal, show that $b^2=4ac$. - Roots equal means discriminant $\Delta=b^2-4ac=0$. - Therefore, $b^2=4ac$. 5. Use the factor theorem to show that $x=2$ is a root of $f(x)=x^3-2x^2+x-2$. - Factor theorem states if $f(k)=0$, then $x-k$ is a factor. - Calculate $f(2)=2^3-2\times2^2+2-2=8-8+2-2=0$. - Since $f(2)=0$, $x=2$ is a root. 6. Factorize $x^3-6x^2+11x-6$ completely. - Try possible roots using factor theorem: factors of 6 are $\pm1, \pm2, \pm3, \pm6$. - Test $x=1$: $1-6+11-6=0$, so $x=1$ is a root. - Divide polynomial by $(x-1)$ to get quadratic: $x^2-5x+6$. - Factor quadratic: $(x-2)(x-3)$. - Complete factorization: $(x-1)(x-2)(x-3)$. 7. Divide $x^3+2x^2-5x+6$ by $x+2$. - Use polynomial division or synthetic division. - Synthetic division with root $-2$: Coefficients: 1, 2, -5, 6 Multiply and add steps: Bring down 1 $1\times(-2)=-2$, $2+(-2)=0$ $0\times(-2)=0$, $-5+0=-5$ $-5\times(-2)=10$, $6+10=16$ - Remainder is 16, quotient is $x^2+0x-5 = x^2 -5$. - So, $\frac{x^3+2x^2-5x+6}{x+2} = x^2 -5 + \frac{16}{x+2}$. 8. If $f(x)=x^3+ax^2+bx+c$ has a root at $x=-1$, show that $f(-1)=0$. - By definition of root, $f(-1)=0$. 9. A die is thrown once. Find the probability of getting an odd number. - Odd numbers on a die: 1, 3, 5 (3 outcomes). - Total outcomes: 6. - Probability: $\frac{3}{6}=\frac{1}{2}$. 10. A bag contains 3 red and 2 blue balls. One ball is picked at random. Find the probability it is red. - Total balls: 3+2=5. - Red balls: 3. - Probability: $\frac{3}{5}$. 11. Two coins are tossed. Find the probability of getting at least one head. - Total outcomes: 4 (HH, HT, TH, TT). - Outcomes with at least one head: 3. - Probability: $\frac{3}{4}$. 12. A card is drawn from a pack of 52 cards. Find the probability of drawing: (i) a king (ii) a spade - (i) Kings: 4 cards. - Probability: $\frac{4}{52}=\frac{1}{13}$. - (ii) Spades: 13 cards. - Probability: $\frac{13}{52}=\frac{1}{4}$. 13. Solve $x^2-4x-5=0$ and use the roots to form a quadratic equation with sum of roots = 6 and product = 5. - Solve original: $x^2-4x-5=0$. - Factor: $(x-5)(x+1)=0$. - Roots: 5 and -1. - New quadratic with sum $S=6$ and product $P=5$ is $x^2 - Sx + P = 0$. - So, $x^2 - 6x + 5=0$. 14. Factorize $x^3-3x^2-4x+12$. - Try roots: factors of 12 are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$. - Test $x=2$: $8-12-8+12=0$, root found. - Divide by $(x-2)$: quotient $x^2 - x - 6$. - Factor quadratic: $(x-3)(x+2)$. - Complete factorization: $(x-2)(x-3)(x+2)$. 15. A box contains 5 white and 3 black balls. Two balls are drawn at random. Find the probability that: (i) both are white (ii) one is white and one is black - Total balls: 8. - Total ways to choose 2 balls: $\binom{8}{2}=28$. - (i) Both white: $\binom{5}{2}=10$. - Probability: $\frac{10}{28}=\frac{5}{14}$. - (ii) One white and one black: $\binom{5}{1}\times\binom{3}{1}=5\times3=15$. - Probability: $\frac{15}{28}$. 16. Solve the system: $2x+y=7$, and $x-y=1$. - Add equations: $(2x+y)+(x-y)=7+1 \Rightarrow 3x=8 \Rightarrow x=\frac{8}{3}$. - Substitute $x$ into $x-y=1$: $\frac{8}{3}-y=1 \Rightarrow y=\frac{8}{3}-1=\frac{5}{3}$. 17. Solve: $3x+2y=12$ and $2x-y=1$. - From second: $y=2x-1$. - Substitute into first: $3x+2(2x-1)=12 \Rightarrow 3x+4x-2=12 \Rightarrow 7x=14 \Rightarrow x=2$. - Then $y=2(2)-1=3$. 18. If $x+y=10$ and $x-y=4$, find $x$ and $y$. - Add: $2x=14 \Rightarrow x=7$. - Substitute: $7+y=10 \Rightarrow y=3$. 19. Solve the simultaneous equations: $x^2+y=10$ and $x+y=6$. - From second: $y=6 - x$. - Substitute into first: $x^2 + (6 - x) = 10 \Rightarrow x^2 - x + 6 = 10 \Rightarrow x^2 - x - 4 = 0$. - Solve quadratic: $x=\frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}$. - Corresponding $y=6 - x$. Final answers: 1. $x=3,4$ 2. $x=1,-2.5$ 3. $x=-2$ (equal roots) 4. $b^2=4ac$ 5. $x=2$ is root 6. $(x-1)(x-2)(x-3)$ 7. Quotient $x^2-5$, remainder 16 8. $f(-1)=0$ 9. $\frac{1}{2}$ 10. $\frac{3}{5}$ 11. $\frac{3}{4}$ 12. (i) $\frac{1}{13}$ (ii) $\frac{1}{4}$ 13. New quadratic: $x^2-6x+5=0$ 14. $(x-2)(x-3)(x+2)$ 15. (i) $\frac{5}{14}$ (ii) $\frac{15}{28}$ 16. $x=\frac{8}{3}, y=\frac{5}{3}$ 17. $x=2, y=3$ 18. $x=7, y=3$ 19. $x=\frac{1 \pm \sqrt{17}}{2}, y=6 - x$