Find the roots of the following by factorisation: (i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0 (iii) 2 2 7 5 2 0 x x (iv) 2x2 – x + 1 8 = 0 (v) 100x2 – 20x + 1 = 0
Step-by-Step Solution
Key Concept: Use the factorisation method (also called the AC method) for a quadratic equation $ax^2+bx+c=0$. Find two numbers whose product is $a\times c$ and whose sum is $b$. Rewrite the middle term using these numbers, factor by grouping, and set each linear factor to zero to obtain the roots.
### (i) $x^{2}-3x-10=0$
1. $a=1,\;b=-3,\;c=-10$; $a\times c = -10$.
2. Numbers whose product $-10$ and sum $-3$ are $-5$ and $2$.
3. Rewrite: $x^{2}-5x+2x-10=0$.
4. Factor by grouping: $x(x-5)+2(x-5)=0$.
5. $(x-5)(x+2)=0$.
6. Roots: $x=5$ or $x=-2$.
### (ii) $2x^{2}+x-6=0$
1. $a=2,\;b=1,\;c=-6$; $a\times c = -12$.
2. Numbers whose product $-12$ and sum $1$ are $4$ and $-3$.
3. Rewrite: $2x^{2}+4x-3x-6=0$.
4. Group: $2x(x+2)-3(x+2)=0$.
5. $(x+2)(2x-3)=0$.
6. Roots: $x=-2$ or $x=\frac{3}{2}$.
### (iii) $2x^{2}+7x+5=0$ *(interpreted from the given notation)*
1. $a=2,\;b=7,\;c=5$; $a\times c = 10$.
2. Numbers whose product $10$ and sum $7$ are $5$ and $2$.
3. Rewrite: $2x^{2}+5x+2x+5=0$.
4. Group: $x(2x+5)+1(2x+5)=0$.
5. $(2x+5)(x+1)=0$.
6. Roots: $x=-\frac{5}{2}$ or $x=-1$.
### (iv) $2x^{2}-x+\frac{1}{8}=0$
1. Multiply by $8$ to clear the fraction: $16x^{2}-8x+1=0$.
2. Recognise a perfect square: $(4x-1)^{2}=16x^{2}-8x+1$.
3. Hence $(4x-1)^{2}=0$.
4. Root (double root): $4x-1=0 \Rightarrow x=\frac{1}{4}$.
### (v) $100x^{2}-20x+1=0$
1. Observe that $(10x-1)^{2}=100x^{2}-20x+1$.
2. Hence $(10x-1)^{2}=0$.
3. Root (double root): $10x-1=0 \Rightarrow x=\frac{1}{10}$.
Summary of roots
- (i) $x=5,\;x=-2$
- (ii) $x=-2,\;x=\frac{3}{2}$
- (iii) $x=-\frac{5}{2},\;x=-1$
- (iv) $x=\frac{1}{4}$ (double root)
- (v) $x=\frac{1}{10}$ (double root)
Correct Answer: (i) $x=5$ and $x=-2$; (ii) $x=-2$ and $x=\frac{3}{2}$; (iii) $x=-\frac{5}{2}$ and $x=-1$; (iv) $x=\frac{1}{4}$ (double root); (v) $x=\frac{1}{10}$ (double root)