JEE Main 2020 - 2 Sep (Evening)

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Animated Solution for Mathematics - Quadratic Equations: Let $f (x)$ be a quadratic polynomial such that $f (- 1) + f (2) = 0$ . If one of the roots of $f (x) = 0$ is 3, then its other root lies in :

Q: Wait, why can we assume $f(x) = a(x - 3)(x - k)$? What if the roots are the same?

That is a fantastic question! The form $f(x) = a(x - r_1)(x - r_2)$ is the general factored form for any quadratic polynomial with roots $r_1$ and $r_2$. If the roots were the same, $k$ would simply equal $3$, and the expression would become $a(x - 3)^2$. Our algebraic approach covers this case automatically, so you do not need to worry about it separately!

Q: Can we just set $a = 1$ to make the math easier?

Yes, you absolutely can! Since $a$ is a non-zero constant, it will appear in every term of the equation $f(-1) + f(2) = 0$. When you set the sum to zero, $a$ will factor out and divide away, meaning the specific value of $a$ does not affect the location of the roots. Setting $a = 1$ is a great shortcut, but keeping it as a variable is a good habit to ensure you don't accidentally lose it in more complex problems.

Q: Why does the problem ask for an interval instead of an exact value?

In many JEE problems, finding the exact value is just the first step. The question asks for the interval to test your ability to place the root on the number line. Once you find $k = -0.4$, you must identify which of the given options contains this value. It is a way to ensure you understand the relationship between the algebraic result and the coordinate geometry.

Select Answer:

Visualized Solution

Identify the Known Root

Given a quadratic polynomial $f (x)$ .
One of the roots is $x = 3$ .

Factored Form of $f (x)$

Since $3$ is a root, $(x - 3)$ is a factor.
Let the unknown second root be $k$ .
We can write: $f (x) = a (x - 3) (x - k)$ , where $a \neq = 0$ .

Evaluate $f (- 1)$

We are given the condition: $f (- 1) + f (2) = 0$ .
First, substitute $x = - 1$ into our assumed $f (x)$ .
$f (- 1) = a (- 1 - 3) (- 1 - k)$

Simplify $f (- 1)$

$f (- 1) = a (- 4) (- 1 - k)$
Factor out $- 1$ from $(- 1 - k)$ :
$f (- 1) = 4 a (1 + k)$

Evaluate $f (2)$

Next, substitute $x = 2$ into $f (x)$ .
$f (2) = a (2 - 3) (2 - k)$

Simplify $f (2)$

$f (2) = a (- 1) (2 - k)$
Distribute the negative sign:
$f (2) = a (k - 2)$

Apply the Given Condition

Substitute the simplified expressions into $f (- 1) + f (2) = 0$ .
$4 a (1 + k) + a (k - 2) = 0$

Eliminate the Constant $a$

Factor out $a$ :
$a [4 (1 + k) + (k - 2)] = 0$
Since $f (x)$ is a quadratic polynomial, $a \neq = 0$ .
Therefore, $4 (1 + k) + (k - 2) = 0$ .

Expand the Equation

Expand the terms inside the bracket:
$4 + 4 k + k - 2 = 0$

Solve for the Root $k$

Combine like terms:
$5 k + 2 = 0$
Isolate $k$ :
$5 k = - 2 ⟹ k = - \frac{2}{5} = - 0.4$

Identify the Correct Interval

The second root is $k = - 0.4$ .
We need to check which given interval contains $- 0.4$ .
$- 1 < - 0.4 < 0$
Therefore, $k \in (- 1, 0)$ .

00:00 / 00:00

The Sigma Insight: Relation Between Roots and Coefficients

Solution Diagram

Analyzing the Setup

We are given that

f (x)

is a quadratic polynomial with one root at

x = 3

. According to the Factor Theorem,

(x - 3)

must be a factor of the polynomial.

Since

f (x)

is a quadratic, it must possess exactly two roots. Let the unknown second root be denoted by

k

We can express the polynomial in its factored form as:

f (x) = a (x - 3) (x - k)

Here,

a

is a non-zero constant representing the leading coefficient of the parabola.

The Power of the Condition

The problem provides the constraint

f (- 1) + f (2) = 0

. To utilize this, we evaluate the function at the given points.

For

x = - 1

f (- 1) = a (- 1 - 3) (- 1 - k) = a (- 4) (- 1 - k) = 4 a (1 + k)

For

x = 2

f (2) = a (2 - 3) (2 - k) = a (- 1) (2 - k) = a (k - 2)

The Algebraic Symphony

We now substitute these expressions into the given condition

f (- 1) + f (2) = 0

4 a (1 + k) + a (k - 2) = 0

Since $a eq 0$, we can divide the entire equation by

a

4 (1 + k) + (k - 2) = 0

Expanding the terms, we obtain:

4 + 4 k + k - 2 = 0

5 k + 2 = 0

Solving for the unknown root

k

k = - \frac{2}{5} = - 0.4

Final Placement

We have determined that the second root of the quadratic polynomial is

k = - 0.4

On the real number line, the value

- 0.4

is located between

- 1

and

0

. Therefore, the root lies in the interval $(- 1, 0)$ .

Common Traps & FAQs

Wait, why can we assume

f (x) = a (x - 3) (x - k)

? What if the roots are the same?

That is a fantastic question! The form

f (x) = a (x - r_{1}) (x - r_{2})

is the general factored form for any quadratic polynomial with roots

r_{1}

and

r_{2}

. If the roots were the same,

k

would simply equal

3

, and the expression would become

a (x - 3)^{2}

. Our algebraic approach covers this case automatically, so you do not need to worry about it separately!

Can we just set

a = 1

to make the math easier?

Yes, you absolutely can! Since

a

is a non-zero constant, it will appear in every term of the equation

f (- 1) + f (2) = 0

. When you set the sum to zero,

a

will factor out and divide away, meaning the specific value of

a

does not affect the location of the roots. Setting

a = 1

is a great shortcut, but keeping it as a variable is a good habit to ensure you don't accidentally lose it in more complex problems.

Why does the problem ask for an interval instead of an exact value?

In many JEE problems, finding the exact value is just the first step. The question asks for the interval to test your ability to place the root on the number line. Once you find

k = - 0.4

, you must identify which of the given options contains this value. It is a way to ensure you understand the relationship between the algebraic result and the coordinate geometry.

Animated Solution for Mathematics - Quadratic Equations: Let $f (x)$ be a quadratic polynomial such that $f (- 1) + f (2) = 0$ . If one of the roots of $f (x) = 0$ is 3, then its other root lies in :

Select Answer:

Visualized Solution

Identify the Known Root

Factored Form of $f (x)$

Evaluate $f (- 1)$

Simplify $f (- 1)$

Evaluate $f (2)$

Simplify $f (2)$

Apply the Given Condition

Eliminate the Constant $a$

Expand the Equation

Solve for the Root $k$

Identify the Correct Interval

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of the Condition

The Algebraic Symphony

Final Placement

Similar Questions

Let $f (x)$ be a quadratic polynomial such that $f (- 2) + f (3) = 0$ . If one of the roots of $f (x) = 0$ is $- 1$ , then the sum of the roots of $f (x) = 0$ is equal to:

If f(x) is a quadratic expression such that $f (1) + f (2) = 0$ , and -1 is a root of $f (x) = 0$ , then the other root of $f (x) = 0$ is :-

If $p$ and $q$ are the roots of the equation $x^{2} + p x + q = 0$ , then

The value of 'a' for which one root of the quadratic equation $(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0$ is twice as large as the other is

If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

If the difference between the roots of the equation $x^{2} + a x + 1 = 0$ is less than $5$ , then the set of possible values of $a$ is

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ , if one of the root is square of the other, then $p$ is equal to

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b}$ , is :

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Let $f (x)$ be a quadratic polynomial with leading coefficient 1 such that $f (0) = p, p \neq = 0$ and $f (1) = \frac{1}{3}$ . If the equation $f (x) = 0$ and $f (f (f (f (x)))) = 0$ have a common real root, then $f (- 3)$ is equal to

Select Answer:

Visualized Solution

Identify the Known Root

Factored Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−1)

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−1)

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

Apply the Given Condition

Eliminate the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Expand the Equation

Solve for the Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Identify the Correct Interval

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of the Condition

The Algebraic Symphony

Final Placement

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a quadratic polynomial such that f(−2)+f(3)=0. If one of the roots of f(x)=0 is −1, then the sum of the roots of f(x)=0 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is a quadratic expression such that f(1)+f(2)=0, and -1 is a root of f(x)=0, then the other root of f(x)=0 is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p and q are the roots of the equation x2+px+q=0, then

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one root is square of the other root of the equation x2+px+q=0, then the relation between p and q is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the difference between the roots of the equation x2+ax+1=0 is less than 5​, then the set of possible values of a is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the equation 3x2+px+3=0,p>0, if one of the root is square of the other, then p is equal to

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Factored Form of $f (x)$

Evaluate $f (- 1)$

Simplify $f (- 1)$

Evaluate $f (2)$

Simplify $f (2)$

Eliminate the Constant $a$

Solve for the Root $k$

Let $f (x)$ be a quadratic polynomial such that $f (- 2) + f (3) = 0$ . If one of the roots of $f (x) = 0$ is $- 1$ , then the sum of the roots of $f (x) = 0$ is equal to:

If f(x) is a quadratic expression such that $f (1) + f (2) = 0$ , and -1 is a root of $f (x) = 0$ , then the other root of $f (x) = 0$ is :-

If $p$ and $q$ are the roots of the equation $x^{2} + p x + q = 0$ , then

The value of 'a' for which one root of the quadratic equation $(a^{2} - 5 a + 3) x^{2} + (3 a - 1) x + 2 = 0$ is twice as large as the other is

If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

If the difference between the roots of the equation $x^{2} + a x + 1 = 0$ is less than $5$ , then the set of possible values of $a$ is

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ , if one of the root is square of the other, then $p$ is equal to

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b}$ , is :

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are