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Q3(iii):
Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases:
(iii) $p(x) = kx^2 – \sqrt{2}x + 1$
Solution :
Initial Setup & Theoretical Foundation
We are given the quadratic polynomial:
$p(x) = kx^2 - \sqrt{2}x + 1$
We are also given a linear binomial, which acts as a divisor:
$g(x) = x - 1$
The objective is to determine the exact value of the unknown coefficient $k$ under the strict condition that $g(x)$ is a factor of $p(x)$.
Step 1: Application of the Factor Theorem
[Per the Factor Theorem of Polynomials], a linear polynomial of the form $x - c$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluates to zero at $x = c$. Mathematically, this is expressed as:
$p(c) = 0$
By comparing our given divisor $g(x) = x - 1$ to the standard form $x - c$, we identify the root of the divisor as:
$x - 1 = 0 \implies x = 1$
Therefore, for $x - 1$ to be a factor of $p(x)$, the polynomial must satisfy the condition:
$p(1) = 0$
Step 2: Substitution and Algebraic Manipulation
We substitute $x = 1$ into the original polynomial $p(x)$ to establish our equation:
- $p(1) = k(1)^2 - \sqrt{2}(1) + 1$
- $p(1) = k(1) - \sqrt{2} + 1$
- $p(1) = k - \sqrt{2} + 1$
Equating this expression to zero [as mandated by the Factor Theorem]:
$k - \sqrt{2} + 1 = 0$
Step 3: Solving for the Unknown Constant $k$
To isolate $k$, we transpose the constant terms to the right side of the equation. When moving terms across the equals sign, their signs invert [by the properties of equality]:
- Add $\sqrt{2}$ to both sides: $k + 1 = \sqrt{2}$
- Subtract $1$ from both sides: $k = \sqrt{2} - 1$
Graphical Verification & Geometric Interpretation
Substituting $k = \sqrt{2} - 1$ back into the original equation yields the specific polynomial:
$p(x) = (\sqrt{2} - 1)x^2 - \sqrt{2}x + 1$
Geometrically, the fact that $x - 1$ is a factor means that the parabola represented by $y = p(x)$ must intersect the x-axis exactly at the coordinate $(1, 0)$. The graph below plots this precise quadratic function, visually confirming the root at $x = 1$.
Final Solution: The value of $k$ that makes $x - 1$ a factor of the polynomial $p(x) = kx^2 - \sqrt{2}x + 1$ is $k = \sqrt{2} - 1$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.3
- Q1(i): Determine which of the following polynomials has $(x + 1)$ a factor : (i) $x^3 + x^2 + x + 1$
- Q1(ii): Determine which of the following polynomials has $(x + 1)$ a factor : (ii) $x^4 + x^3 + x^2 + x + 1$
- Q1(iii): Determine which of the following polynomials has $(x + 1)$ a factor : (iii) $x^4 + 3x^3 + 3x^2 + x + 1$
- Q1(iv): Determine which of the following polynomials has $(x + 1)$ a factor : (iv) $x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}$
- Q2(i): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (i) $p(x) = 2x^3 + x^2 – 2x – 1, g(x) = x + 1$
- Q2(ii): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$
- Q2(iii): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = x^3 – 4x^2 + x + 6, g(x) = x – 3$
- Q3(i): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (i) $p(x) = x^2 + x + k$
- Q3(ii): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = 2x^2 + kx + \sqrt{2}$
- Q3(iv): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (iv) $p(x) = kx^2 – 3x + k$
- Q4(i): Factorise : (i) $12x^2 – 7x + 1$
- Q4(ii): Factorise : (ii) $2x^2 + 7x + 3$
- Q4(iii): Factorise : (iii) $6x^2 + 5x – 6$
- Q4(iv): Factorise : (iv) $3x^2 – x – 4$
- Q5(i): Factorise : (i) $x^3 – 2x^2 – x + 2$
- Q5(ii): Factorise : (ii) $x^3 – 3x^2 – 9x – 5$
- Q5(iii): Factorise : (iii) $x^3 + 13x^2 + 32x + 20$
- Q5(iv): Factorise : (iv) $2y^3 + y^2 – 2y – 1$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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