Question

1.if one zero of the polynomial (a²+9)x²+13x+6a is reciprocal of the other,find the value of a..​

Answer 1

$\large\sf\underline{Given\::}$

Polynomial :

• $\sf\:(a^{2}+9) x^{2}+13x+6a$

Condition given is :

• One zero of the given polynomial is reciprocal of the other.

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$\large\sf\underline{To\:find\::}$

• The value of a in the polynomial

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$\large\sf\underline{Assumption\::}$

Let one root of the polynomial be $\sf\:\alpha$ .

Then the other root would be reciprocal of $\sf\:\alpha$ i.e., $\bf\:\frac{1}{\alpha}$ .

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$\large\sf\underline{Solution\::}$

We know that :

$\spadesuit$ $\sf\large\color{purple}{Product\:of\:the\:root\:=\frac{c}{a}}$

So now we would frame an equation using the above relation.

• Product of the root here means $\sf\:\alpha \times \frac{1}{\alpha}$

Now how would we find c and a . For finding it's value we would compare the given polynomial $\sf\:(a^{2}+9) x^{2}+13x+6a$ with $\sf\:ax^{2}+bx+c=0$ .

Comparing we get :

• a = $\sf\:(a^{2}+9)$

• c = $\sf\:6a$

So now let's quickly frame the equation :

$\sf\twoheadrightarrow\:\alpha \times \frac{1}{\alpha}=\frac{6a}{(a^{2}+9)}$

• Multiplying the terms in LHS

$\sf\twoheadrightarrow\:\frac{\alpha}{\alpha}=\frac{6a}{(a^{2}+9)}$

• Reducing the fraction in LHS to lower terms

$\sf\twoheadrightarrow\:\cancel{\frac{\alpha}{\alpha}}=\frac{6a}{(a^{2}+9)}$

$\sf\twoheadrightarrow\:1=\frac{6a}{(a^{2}+9)}$

• Now cross multiplying

$\sf\twoheadrightarrow\:a^{2}+9=6a$

• Transposing +6a to other side , it becomes -6a

$\sf\twoheadrightarrow\:a^{2}+9-6a=0$

$\sf\twoheadrightarrow\:a^{2}-6a+9=0$

We got an quadratic equation so let's factorise it using middle term breaking method.

• Middle term i.e., 6a can be splitted as -3a and -3a

$\sf\twoheadrightarrow\:a^{2}-(3+3)a+9=0$

$\sf\twoheadrightarrow\:a^{2}-3a-3a+9=0$

• Taking a and 3 common from first two terms and last two terms

$\sf\twoheadrightarrow\:a(a-3)-3(a-3)=0$

• Taking (a - 3) as common from whole terms

$\sf\twoheadrightarrow\:(a-3)(a-3)=0$

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$\sf\therefore\:Value\:of\:a\:$

$\sf\leadsto\:(a-3) =0$

$\sf\leadsto\:a-3=0$

$\small{\underline{\boxed{\mathrm\red{\leadsto\:a\:=\:3}}}}$ ★

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!! Hope it helps !!

Answer 2

Answer:

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