Question

Question 16
Quadratic polynomial 4x^2+12x+9 has zeroes as p and q Now forma a quadratic polynomial whose
zeroes are p-1 and q-1​

Answer 1

Given :

• A quadratic polynomial 4x² + 12x + 9 has zeroes p and q

To Find :

• The quadratic polynomial whose zeroes are p - 1 and q - 1.

Knowldege required :

The qudratic equation when its roots α , β are given is given by ,

$\\ : \implies \sf \: x^2-(\alpha + \beta)x+\alpha\beta$

Solution :

First of all let us find the zeroes of 4x² +12x + 9.

$\\ : \implies \sf \: 4 {x}^{2} + 12x + 9 = 0$

$\\ :\implies \sf \: 4 {x}^{2} + 6x + 6x + 9 = 0$

$\\ : \implies \sf \: 2x(2x + 3) + 3(2x + 3) = 0$

$\\ : \implies \sf \: (2x + 3)(2x + 3) = 0$

$\\ : \implies \sf \: x = - \frac{3}{2}$

Since we are given that the zeroes as p , q . Then p and q must be ,

$\\ : \implies \sf \: p = - \frac{3}{2} \: and \: q = - \frac{3}{2}$

We are given that other quadratic equation has p - 1 and q - 1 as roots. Then the values of roots of that quadratic eqiation are ,

$\\ : \implies \sf \: p - 1 = \frac{ - 3}{2} - 1$

$\\ : \implies \sf \: p - 1 = \frac{ - 5}{2} \:$

The value of q - 1 is same as p - 1. Since p = q.

$\\ : \implies \sf \: q - 1 = - \frac{5}{2}$

Now , by applying the above condition . Sum of the roots is

$\\ : \implies \sf \:( p - 1)+( q - 1) = \frac{ - 5}{2} + \frac{ - 5}{2}$

$\\ : \implies \sf \: (p - 1) + (q - 1) = \frac{ - 10}{2} = -5$

Now the value of product of roots is ;

$\\ : \implies \sf \: (p - 1)(q - 1)= \frac{ - 5}{2} \times \frac{ - 5}{2}$

$\\ : \implies \sf \: (p - 1)(q - 1) = \frac{25}{4}$

Then the required quadratic equation becomes ,

$\\ : \implies \sf \: {x}^{2} -(-5)x + \dfrac{25}{4}=0$

Now multiplying the whole equation with 4 we get ;

$\\ :\implies\underline{\boxed{\sf\:{4x^2+20x+25 =0}}}$

Hence , The Required quadratic equation is 4x² + 20x + 25

Answer 2

Given :

• 4x² + 12x + 9

To Find :

• Now forma a quadratic polynomial whose zeroes are p - 1 and q - 1

Solution :

Therefore p + q = -12/4= -3

p × q = c/a = 9/4

Now quad eqn of roots(p - 1) and ( q - 1)

= x² -( p - 1+ q - 1)x + (p - 1).(q - 1 ) = 0

= x² -(p + q - 2)x+ (p × q -(p + q ) + 1 ) = 0

= x² -(- 3 - 2 )x+ (9/4 + 3 + 1 ) = 0

= x² + 5x + (9 + 12 + 4)/4 = 0

= 4x² + 20x + 25 = 0