3 Marks Question

Question 13 Marks

Solve for $ x : \frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3}, x \neq 1,2,3$

Answer

Solving for $x$, using common denominator
$\Rightarrow \frac{x-3+x-1}{(x-1)(x-2)(x-3)}=\frac{2}{3}$
$\Rightarrow \frac{2 x-4}{(x-1)(x-2)(x-3)}=\frac{2}{3}$
Cancel out like terms
$\Rightarrow \frac{2(x-2)}{(x-1)(x-2)(x-3)}=\frac{2}{3}$
Cross multiply
$\Rightarrow 3=(x-1)(x-3)$
$\Rightarrow x^2-3 x-x+3=3$
$\Rightarrow x^2-4 x=0$
$\Rightarrow x(x-4)=0$
Hence, $x=0$ and $x-4=0$
Therefore, $x=0$ and $x=4$.

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Question 23 Marks

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.
$2 x^2-3+5 x$

Answer

Let $p(x)=2 x^2+5 x-3$
Zero of the polynomial is the value of $x$ where $p(x)=0$. So, equate the given expression with $0$ , we get:
$2 x^2+5 x-3=0$
Compare the given equation with the general equation $a x^2+b x+c$, we get:
$a=2, b=5, c=-3$
Let, $\alpha$ and $\beta$ ne the roots of the given equation.
Find the roots of the given equation by using quadratic formula:
$\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-5 \pm \sqrt{25-4 \times 2 \times(-3)}}{2 \times 2}$
$\frac{-5 \pm \sqrt{25+24}}{4}=\frac{-5 \pm \sqrt{49}}{4}=\frac{-5 \pm 7}{4}$
So, the roots will be:
$\alpha=\frac{-5+7}{4}=\frac{2}{4}=\frac{1}{2}$
$\beta=\frac{-5-7}{4}=\frac{-12}{4}=-3$
Therefore, $\alpha=\frac{1}{2}$ and $\beta=-3$ are the roots/zeroes of the given equation.
Now, we verify the relationship between the zeroes and the coefficients. We know that:
$\text { Sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}$
i.e. $\alpha+\beta=-\frac{b}{a}$
$\text{LHS}$
$\alpha+\beta=\frac{1}{2}-3=-\frac{5}{2}$
$\text{RHS}$
$-\frac{b}{a}=-\frac{5}{2}$
Therefore, $\text{LHS} = \text{RHS}$
$\text { Product of zeroes }=\frac{\text { Constant term }}{\text { Coefficient of } x^2}$
$\Rightarrow \alpha \times \beta=\frac{c}{\alpha}$
$\text{LHS}$
$\frac{1}{2} \times-3=-\frac{3}{2}$
$\text { RHS }$
$\frac{\text { Constant term }}{\text { Coefficient of } x^2}=\frac{-3}{2}=-\frac{3}{2}$
$\text { Therefore, LHS }=\text { RHS. }$
As in both the conditions, $\text{LHS} = \text{RHS}$, therefore the relationship between coefficient and the zeroes is verified.

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