Question

Quadratic polynomial 4x^2+12x+9 has zeroes as p and q. Now form a Quadratic polynomial whose zeroes are p-1 and q-1​

Answer 1

Step-by-step explanation:

Given:-

• A quadratic polynomial 4x² + 12x + 9

• The zeroes of the polynomial are p and q.

To Find:-

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

Concept used:-

For a quadratic polynomial ax² + bx + c

Sum of zeroes = $\sf \dfrac{-b}{a}$

Product of zeroes = $\sf \dfrac{c}{a}$

Solution:-

Comparing 4x² + 12x + 9 with ax² + bx + c

Here:-

• a = 4 $\:\:\:\:$ • b = 12 $\:\:\:\:$ • c = 9

$\sf Sum \: of \: zeroes = \dfrac{-12}{4} = -3$

$\implies \bf p + q = -3$

$\sf Product \: of \: zeroes = \dfrac{9}{4}$

$\implies \bf pq = \dfrac{9}{4}$

Now, we need to find the quadratic equation whose zeroes are p - 1 and q - 1

Sum of zeroes:-

= p - 1 + q - 1

= (p + q) - 2

= - 3 - 2 = -5

∴ Sum of roots = - 5

Product of zeroes:-

= ( p - 1)(q - 1)

= pq - p - q + 1

= pq - (p + q) + 1

= $\dfrac{9}{4} - (-3) + 1$

= $\dfrac{9}{4} + 4$

= $\dfrac{25}{4}$

∴ Product of zeroes = $\dfrac{25}{4}$

The general form of a quadratic equation is:-

$\sf x^2 - (sum\: of \: zeroes)x + product\: of \: zeroes = 0$

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

$\sf = 4x^2 + 20x + 25 = 0$

Therefore:-

$\underline{\boxed{\sf Required \: quadratic \: polynomial \: is \: 4x^2 +20x + 25}}$

Answer 2

⛤Given Question:-

Quadratic polynomial 4x^2+12x+9 has zeroes as p and q. Now form a Quadratic polynomial whose zeroes are p-1 and q-1.

⛤Required Answer:-

The Required quadratic equation is 4x² + 20x + 25.

⛤Explanation:-

$\large \sf \color{green}{Given:- }$

• Quadratic polynomial 4x^2+12x+9 has zeroes as p and q.

$\large \sf \color{green}{To \: Find:- }$

• Form a Quadratic polynomial whose zeroes are p-1 and q-1.

$\large \sf \color{green}{Formula \: used:- }$

➪The qudratic equation when its roots α , β, Formula Used is

$➪ \tt { {x}^{2} } - { \tt{ \alpha + \beta}} + \tt{ \alpha \beta}$

$\large \sf \color{green}{Dealing:- }$

➪Let us find the zeros of 4x² +12x + 9:-

$\leadsto \tt{ {4x}^{2} + 12x + 9 = 0}$

➪By middle term splitting we will solve further:-

$\leadsto \tt{ {4x}^{2} + 6x + 6x + 9 = 0}$

➪Now let's find upto their common factors.

$\leadsto \tt {2x(2x+3)+3(2x+3)=0}$

$\leadsto \tt{(2x+3)+(2x+3)=0}$

➪(2x+3) is common, so let's move further.

$\leadsto \tt {2x=0 - 3}$

$\leadsto \tt {2x=- 3}$

$\leadsto \tt {x= \frac{ - 3}{2} }$

➪So, p and q can be the same, i.e.

$\leadsto \tt {p \: or \: q= \frac{ - 3}{2} }$

➪It is given to form a Quadratic polynomial whose zeroes are p-1 and q-1. So the quadratic equation will be:-

$\leadsto \tt{p - 1= - \frac{3}{2} - 1}$

$\leadsto \tt{p - 1= \frac{3}{2} + 1}$

➪We xan find it shortly by multiplying 1 with 2 and adding 3 to it.

$\leadsto \tt{p - 1= \frac{5}{2} }$

➪As given above, the value of p-1 and q-1 is same, so the value of q-1 will be:-

$\leadsto \tt{q - 1= \frac{5}{2} }$

➪Now let's apply all the values:-

$\leadsto \tt{p - 1 \times q - 1= \frac{5}{2} \times \frac{5}{2} }$

$\leadsto \tt{p - 1 \times q - 1= \frac{25}{4} }$

➪So, let's find the result:-

$\leadsto \tt{ {x}^{2} + 5x + \frac{25}{4} = 0 }$

➪By multiplying it by 4, we will get

$\leadsto \tt{ ({x}^{2} + 5x + \frac{25}{4}) \times 4 = 0 }$

$\leadsto \tt{ 4 {x}^{2} + 20x + 25 = 0 }$

➪So the required Quadratic equation is

${\bigstar}\huge \pink{ \boxed{\tt{ 4 {x}^{2} + 20x + 25 }}}$

__________________