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Q3(vii):
Verify whether the following are zeroes of the polynomial, indicated against them.
(vii) $p(x) = 3x^2 – 1, x = –\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
Solution :
Initial Setup & Theoretical Foundation
We are given the quadratic polynomial:
$p(x) = 3x^2 - 1$
We must verify whether the given values, $x = -\frac{1}{\sqrt{3}}$ and $x = \frac{2}{\sqrt{3}}$, are zeroes of the polynomial.
[Theoretical Justification: By the definition of the zero of a polynomial, a real number $a$ is a zero of a polynomial $p(x)$ if and only if $p(a) = 0$. Therefore, we must substitute each given value of $x$ into the polynomial and evaluate the result.]
Step 1: Evaluating the Polynomial at $x = -\frac{1}{\sqrt{3}}$
Substitute $x = -\frac{1}{\sqrt{3}}$ into the polynomial $p(x)$:
$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1$
First, apply the exponent to the fraction. The square of a negative number is positive, and the square of a square root removes the radical:
$\left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{(-1)^2}{(\sqrt{3})^2} = \frac{1}{3}$
Now, substitute this back into the equation:
$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(\frac{1}{3}\right) - 1$
$p\left(-\frac{1}{\sqrt{3}}\right) = 1 - 1 = 0$
Since $p\left(-\frac{1}{\sqrt{3}}\right) = 0$, the value $x = -\frac{1}{\sqrt{3}}$ satisfies the condition.
Step 2: Evaluating the Polynomial at $x = \frac{2}{\sqrt{3}}$
Substitute $x = \frac{2}{\sqrt{3}}$ into the polynomial $p(x)$:
$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1$
Apply the exponent to the numerator and the denominator:
$\left(\frac{2}{\sqrt{3}}\right)^2 = \frac{2^2}{(\sqrt{3})^2} = \frac{4}{3}$
Substitute this back into the equation:
$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{4}{3}\right) - 1$
$p\left(\frac{2}{\sqrt{3}}\right) = 4 - 1 = 3$
Since $p\left(\frac{2}{\sqrt{3}}\right) = 3 \neq 0$, the value $x = \frac{2}{\sqrt{3}}$ does not satisfy the condition.
Graphical Analysis
To visualize this algebraically proven result, we can observe the graph of the parabola $y = 3x^2 - 1$. The zeroes of the polynomial correspond to the $x$-intercepts (where the graph crosses the horizontal axis, $y=0$).
As demonstrated by the graph, the point $x = -\frac{1}{\sqrt{3}}$ lies exactly on the $x$-axis (making it a zero), whereas the point $x = \frac{2}{\sqrt{3}}$ maps to a $y$-value of $3$, confirming it is not a zero.
Final Conclusion
Final Solution: $x = -\frac{1}{\sqrt{3}}$ is a zero of the polynomial $p(x) = 3x^2 - 1$, whereas $x = \frac{2}{\sqrt{3}}$ is not a zero of the polynomial.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2
- Q1(i): Find the value of the polynomial $5x – 4x^2 + 3$ at (i) $x = 0$
- Q1(ii): Find the value of the polynomial $5x – 4x^2 + 3$ at (ii) $x = –1$
- Q1(iii): Find the value of the polynomial $5x – 4x^2 + 3$ at (iii) $x = 2$
- Q2(i): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (i) $p(y) = y^2 – y + 1$
- Q2(ii): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (ii) $p(t) = 2 + t + 2t^2 – t^3$
- Q2(iii): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iii) $p(x) = x^3$
- Q2(iv): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iv) $p(x) = (x – 1) (x + 1)$
- Q3(i): Verify whether the following are zeroes of the polynomial, indicated against them. (i) $p(x) = 3x + 1, x = –\frac{1}{3}$
- Q3(ii): Verify whether the following are zeroes of the polynomial, indicated against them. (ii) $p(x) = 5x – \pi, x = \frac{4}{5}$
- Q3(iii): Verify whether the following are zeroes of the polynomial, indicated against them. (iii) $p(x) = x^2 – 1, x = 1, –1$
- Q3(iv): Verify whether the following are zeroes of the polynomial, indicated against them. (iv) $p(x) = (x + 1) (x – 2), x = – 1, 2$
- Q3(v): Verify whether the following are zeroes of the polynomial, indicated against them. (v) $p(x) = x^2, x = 0$
- Q3(vi): Verify whether the following are zeroes of the polynomial, indicated against them. (vi) $p(x) = lx + m, x = –\frac{m}{l}$
- Q3(viii): Verify whether the following are zeroes of the polynomial, indicated against them. (viii) $p(x) = 2x + 1, x = \frac{1}{2}$
- Q4(i): Find the zero of the polynomial in each of the following cases: (i) $p(x) = x + 5$
- Q4(ii): Find the zero of the polynomial in each of the following cases: (ii) $p(x) = x – 5$
- Q4(iii): Find the zero of the polynomial in each of the following cases: (iii) $p(x) = 2x + 5$
- Q4(iv): Find the zero of the polynomial in each of the following cases: (iv) $p(x) = 3x – 2$
- Q4(v): Find the zero of the polynomial in each of the following cases: (v) $p(x) = 3x$
- Q4(vi): Find the zero of the polynomial in each of the following cases: (vi) $p(x) = ax, a \neq 0$
- Q4(vii): Find the zero of the polynomial in each of the following cases: (vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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