Let $p(x)=x^2+b x+c$ be quadratic polynomial with real coefficients $b$ and $c$. Suppose $p(1)=5$ an...

To solve for the product of the roots of the quadratic polynomial $ p(x) = x^2 + bx + c $, given the conditions $ p(1) = 5 $ and $ p(-1) = 3 $:Firstly, use the conditions provided:$ p(1) = 1 + b + c = 5 $$ p(-1) = 1 - b + c = 3 $Add the two equations:$ (1 + b + c) + (1 - b + c) = 5 + 3 $This simplifies to:$ 2 + 2c = 8 $Subtract 2 from both sides:$ 2c = 6 $Divide by 2:$ c = 3 $The product of the roots of the quadratic equation $ p(x) = x^2 + bx + c = 0 $ is given by:$ \alpha \beta = \frac{c}{1} = c $So, the product of the roots is:$ \alpha \beta = 3 $

Quadratic Equations - IAT-IISER Maths PYQs

To solve for the product of the roots of the quadratic polynomial

p(x)%20%3D%20x%5E2%20%2B%20bx%20%2B%20c

, given the conditions

p(1)%20%3D%205

and

p(-1)%20%3D%203

:Firstly, use the conditions provided:

p(1)%20%3D%201%20%2B%20b%20%2B%20c%20%3D%205

p(-1)%20%3D%201%20-%20b%20%2B%20c%20%3D%203

Add the two equations:

(1%20%2B%20b%20%2B%20c)%20%2B%20(1%20-%20b%20%2B%20c)%20%3D%205%20%2B%203

This simplifies to:

2%20%2B%202c%20%3D%208

Subtract 2 from both sides:

2c%20%3D%206

Divide by 2:

c%20%3D%203

The product of the roots of the quadratic equation

p(x)%20%3D%20x%5E2%20%2B%20bx%20%2B%20c%20%3D%200

is given by:

%5Calpha%20%5Cbeta%20%3D%20%5Cfrac%7Bc%7D%7B1%7D%20%3D%20c

So, the product of the roots is:

%5Calpha%20%5Cbeta%20%3D%203

Quadratic Equations

High-Yield Trend

Questions
1 MCQs

Official Solution

Quadratic Equations

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs