Question

Please answer , very quickly and correctly with a step by step explanation

Answer 1

❍ We are given a polynomial $\sf (a²+9)x²+13x+6a$ whose zeroes are reciprocal of each other.

❍ we are asked to find the value of a.

_______________________

We know that we have a relationship between zeros and the coefficient of that equation.

For any equation:-

• a = coefficient of x²
• b = coefficient of x
• c = constant

in the given equation:-

• a = $\frak{(a²+9)}$

• b = $\frak{13}$

• c = $\frak{6a}$

❍ Now, Let's Consider the zeros of this equation as :-

• $\sf \alpha$ and $\sf\dfrac{1}{\alpha}$

The relationship between product of zeros and coefficient of equations is as follows:-

$\leadsto \large \underline { \boxed {\pink{{ \frak{ \alpha.\beta = \frac{c}{a} }}}}} \bigstar$

In this question we have assumed ß as 1/alpha

$\: \: \: \: \: \: \: \: \: \: \: \dag \: \underline \frak{substituting \: the \: given \: values \: in \: formula}$

$: \implies \sf \alpha \bigg( \dfrac{1}{ \alpha} \bigg) = \dfrac{ {a }^{2} + 9}{6a}$

$: \implies \sf 1 = \dfrac{ {a}^{2} + 9}{6a}$

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

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

Now, we need to split the middle term in order to find the value of a.

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

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

$\rightarrow \sf a(a - 3) - 3(a - 3)$

$\rightarrow \sf (a - 3)(a - 3)$

$\rightarrow\boxed {\pink{{ \frak{a = 3,3 }}}} \bigstar$

$\therefore \underline{ \sf 3 \: is \: our \: required \: answer}$

Answer 2

Given :-

A Quadratic polynomial i.e ( a² + 9 ) x² + 13x + 6a and the zeroes are reciprocal of each other .

To Find :-

The Value of "a" .

Used Concepts :-

• A general Quadratic equation is in the form of "ax² + bx + c = 0" .
• Discrimanant ( D ) of a quadratic equation is given by b² - 4ac .
• Sum of zeroes of Quadratic equation is " -b/a " and that for product is " c/a " .

Solution :-

Let , p ( x ) = ( a² + 9 ) x² + 13x + 6a

For zeroes p ( x ) = 0 ,

( a² + 9 ) x² + 13x + 6a = 0

Now , you can see that it is in the form of a Quadratic equation . Where ,

a = ( a² + 9 )

b = 13

c = 6a

Now , Let one root = ā

Then , another root = 1/ā

Now , Product of roots = c/a = 6a/a² + 9

ā × 1/ā = 6a/a² + 9

6a/ a² + 9 = 1

=> 6a = a² + 9

=> a² + 9 - 6a = 0 => a² - 6a + 9 = 0

Now , a² - 3a - 3a + 9 = 0

=> a ( a - 3 ) -3 ( a - 3 ) = 0

=> ( a - 3 ) ( a - 3 ) = 0

Therefore , a - 3 = 0 => a = 3

Hence , Our required answer is 3 .

Additional Information :-

• The vertex of a parabola is given by ( -b/2a , -D/4a )
• If the coefficient of x² is negative then , the parabola opens downwards .
• If the coefficient of x² is positive then , the parabola opens upwards .
• If the parabola is in variable "x" and not touching the x - axis anywhere so the roots are imaginary.
• If the parabola is only touching X - axis at one point . So , the zeroes of Quadratic equation are equal and D is 0 .
• Three methods to solve a quadratic equation , But the government deleted one of them i.e Completing the square because of the corona pandemic .

1. Quadratic formulae
2. Factorisation .
3. Completing the square .

• The Quadratic Formula i.e -b + √D/2a , -b - √D / 2a .

Note that there are some equations to whom factorisation and completing the square method didn't works . But the Quadratic formula is applicable to all Quadratic equations .

I hope you will read my full answer .