Question

Let p(x) be a quadratic polynomial with leading
Coefficient 1 such that p(1)=3 and p(2)=4. Find
e value of p(-2)​

Answer 1

The value of p(-2) = 12

Given :

p(x) be a quadratic polynomial with leading Coefficient 1 such that p(1) = 3 and p(2) = 4

To find :

The value of p( - 2)

Solution :

Step 1 of 4 :

Assume the polynomial

Here it is given that p(x) be a quadratic polynomial with leading Coefficient 1

Let us assume that the required quadratic polynomial is

$\sf p(x) = {x}^{2} + ax + b \: \: \: - - - (1)$

Step 2 of 4 :

Find the value of a and b

It is given that p(1) = 3 and p(2) = 4

Now p(1) = 3 gives

$\sf p(1) = {1}^{2} + a.1 + b$

$\displaystyle \sf{ \implies 3 = 1 + a + b}$

$\displaystyle \sf{ \implies a + b =2 \: \: \: - - - (2)}$

Now p(2) = 4 gives

$\sf p(2) = {2}^{2} + a.2 + b$

$\displaystyle \sf{ \implies 4 = 4 + 2a + b}$

$\displaystyle \sf{ \implies 2a + b = 0 \: \: \: - - - (3)}$

Solving Equation 2 and Equation 3 we get

a = - 2 , b = 4

Step 3 of 4 :

Find the polynomial

Putting the value of a and b in Equation 1 we get

$\sf p(x) = {x}^{2} - 2x + 4 \: \: \: - - (4)$

Step 4 of 4 :

Find the value of p( - 2)

Putting x = - 2 in Equation 4 we get

$\displaystyle \sf{p( - 2) = {( - 2)}^{2} - 2.( - 2) + 4 }$

$\displaystyle \sf{ \implies p( - 2) = 4 + 4 + 4 }$

$\displaystyle \sf{ \implies p( - 2) = 12}$

The value of p(-2) = 12