Question

Answer the 10th question!!
Both 2nd subdivision nd 3rd subdivision.

Answer 1

Hey There!!!


First, some concept: 

If an algebraic expression is given as $ax^2+bx+c$, then its zeros can be found by putting $ax^2+bx+c=0$ 

Let's assume zeros are $\alpha$ and $\beta$. 
Then, we have the following relations between zeros and coefficients:

$\alpha+\beta = \frac{-b}{a} \\ \\ \alpha \beta = \frac{c}{a}$

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Now let us come to your questions:
Let's take your first one:
It can be factorised as shown:

$abx^2+(b^2-ac)x-bc \\ \\ =abx^2+b^2x-acx-bc \\ \\ = bx(ax+b)-c(ax+b) \\ \\ =(ax+b)(bx-c)$

For zeros,
$(ax+b)(bx-c)=0 \\ \\ \implies \boxed{x=-\frac{b}{a}}\,\,OR\,\,\boxed{x=\frac{c}{b}}$

Now, we can verify the relation between zeros and co-efficients. 

Sum of zeros is:

$-\frac{b}{a}+\frac{c}{b} = -\frac{b^2-ac}{ab} = -\frac{\textrm{Coefficient of }x}{\textrm{Coefficient of }x^2}$

Also, Product of Zeros is:
$-\frac{b}{a}\frac{c}{b} = -\frac{c}{b} = \frac{\textrm{Constant Term}}{\textrm{Coefficient of }x^2}$

Hence Verified.
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Let's take your second question: 

It can be factorised as shown:

$\sqrt{3}x^2+10x+7\sqrt{3} \\ \\ =\sqrt{3}x^2+3x+7x+7\sqrt{3} \\ \\ = \sqrt{3}x(x+\sqrt{3})+7(x+\sqrt{3}) \\ \\ =(x+\sqrt{3})(\sqrt{3}x+7)$

For zeros,
$(x+\sqrt{3})(\sqrt{3}x+7)=0 \\ \\ \implies \boxed{x=-\sqrt{3}}\,\,OR\,\,\boxed{x=-\frac{7}{\sqrt{3}}}$

Now, we can verify the relation between zeros and co-efficients. 

Sum of zeros is:

$-\sqrt{3}-\frac{7}{\sqrt{3}} = -\frac{10}{\sqrt{3}} = -\frac{\textrm{Coefficient of }x}{\textrm{Coefficient of }x^2}$

Also, Product of Zeros is:
$(-\sqrt{3})\left(-\frac{7}{\sqrt{3}}\right) = 7 = \frac{\textrm{Constant Term}}{\textrm{Coefficient of }x^2}$

Hence Verified.

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Hope it helps
Purva
Brainly Community