PolynomialsClass 10 Mathematics NCERT Solutions

To find the zeroes, we factorize the polynomial: $x^{2}-2 x-8 = x^{2}-4x+2x-8 = x(x-4)+2(x-4) = (x-4)(x+2)$ The value of $p(x)$ is zero when $x-4=0$ or $x+2=0$. So, the zeroes are $4$ and $-2$.

Verification: Let the zeroes be $\alpha = 4$ and $\beta = -2$. For the polynomial $x^{2}-2x-8$, we have $a=1, b=-2, c=-8$. Sum of zeroes: $\alpha + \beta = 4 + (-2) = 2$ From coefficients: $-\frac{b}{a} = -\frac{(-2)}{1} = 2$ Product of zeroes: $\alpha\beta = 4 \times (-2) = -8$ From coefficients: $\frac{c}{a} = \frac{-8}{1} = -8$ Hence, the relationship is verified.

To find the zeroes, we factorize the polynomial: $4s^{2}-4s+1 = (2s)^{2} - 2(2s)(1) + 1^{2} = (2s-1)^{2}$ The value of $p(s)$ is zero when $2s-1=0$. So, the zeroes are $\frac{1}{2}$ and $\frac{1}{2}$.

Verification: Let the zeroes be $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{2}$. For the polynomial $4s^{2}-4s+1$, we have $a=4, b=-4, c=1$. Sum of zeroes: $\alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1$ From coefficients: $-\frac{b}{a} = -\frac{(-4)}{4} = 1$ Product of zeroes: $\alpha\beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ From coefficients: $\frac{c}{a} = \frac{1}{4}$ Hence, the relationship is verified.

First, we write the polynomial in standard form: $6x^{2}-7x-3$. To find the zeroes, we factorize by splitting the middle term: $6x^{2}-7x-3 = 6x^{2}-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$ The value of $p(x)$ is zero when $3x+1=0$ or $2x-3=0$. So, the zeroes are $-\frac{1}{3}$ and $\frac{3}{2}$.

Verification: Let the zeroes be $\alpha = -\frac{1}{3}$ and $\beta = \frac{3}{2}$. For the polynomial $6x^{2}-7x-3$, we have $a=6, b=-7, c=-3$. Sum of zeroes: $\alpha + \beta = -\frac{1}{3} + \frac{3}{2} = \frac{-2+9}{6} = \frac{7}{6}$ From coefficients: $-\frac{b}{a} = -\frac{(-7)}{6} = \frac{7}{6}$ Product of zeroes: $\alpha\beta = -\frac{1}{3} \times \frac{3}{2} = -\frac{1}{2}$ From coefficients: $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$ Hence, the relationship is verified.

To find the zeroes, we factorize the polynomial: $4u^{2}+8u = 4u(u+2)$ The value of $p(u)$ is zero when $4u=0$ or $u+2=0$. So, the zeroes are $0$ and $-2$.

Verification: Let the zeroes be $\alpha = 0$ and $\beta = -2$. For the polynomial $4u^{2}+8u$, we have $a=4, b=8, c=0$. Sum of zeroes: $\alpha + \beta = 0 + (-2) = -2$ From coefficients: $-\frac{b}{a} = -\frac{8}{4} = -2$ Product of zeroes: $\alpha\beta = 0 \times (-2) = 0$ From coefficients: $\frac{c}{a} = \frac{0}{4} = 0$ Hence, the relationship is verified.

To find the zeroes, we use the identity $a^2-b^2=(a-b)(a+b)$: $t^{2}-15 = t^{2} - (\sqrt{15})^{2} = (t-\sqrt{15})(t+\sqrt{15})$ The value of $p(t)$ is zero when $t-\sqrt{15}=0$ or $t+\sqrt{15}=0$. So, the zeroes are $\sqrt{15}$ and $-\sqrt{15}$.

Verification: Let the zeroes be $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$. For the polynomial $t^{2}-15$, we can write it as $t^{2}+0t-15$. So, $a=1, b=0, c=-15$. Sum of zeroes: $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$ From coefficients: $-\frac{b}{a} = -\frac{0}{1} = 0$ Product of zeroes: $\alpha\beta = (\sqrt{15}) \times (-\sqrt{15}) = -15$ From coefficients: $\frac{c}{a} = \frac{-15}{1} = -15$ Hence, the relationship is verified.

To find the zeroes, we factorize by splitting the middle term: $3x^{2}-x-4 = 3x^{2}-4x+3x-4 = x(3x-4)+1(3x-4) = (x+1)(3x-4)$ The value of $p(x)$ is zero when $x+1=0$ or $3x-4=0$. So, the zeroes are $-1$ and $\frac{4}{3}$.

Verification: Let the zeroes be $\alpha = -1$ and $\beta = \frac{4}{3}$. For the polynomial $3x^{2}-x-4$, we have $a=3, b=-1, c=-4$. Sum of zeroes: $\alpha + \beta = -1 + \frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3}$ From coefficients: $-\frac{b}{a} = -\frac{(-1)}{3} = \frac{1}{3}$ Product of zeroes: $\alpha\beta = -1 \times \frac{4}{3} = -\frac{4}{3}$ From coefficients: $\frac{c}{a} = \frac{-4}{3}$ Hence, the relationship is verified.

A quadratic polynomial with zeroes $\alpha$ and $\beta$ can be written in the form $p(x) = k(x^2 - (\alpha+\beta)x + \alpha\beta)$, where $k$ is any non-zero real number. We can choose $k=1$ for simplicity or another value to get integer coefficients.

(i) Sum = $\frac{1}{4}$, Product = $-1$ Let the sum of zeroes $\alpha+\beta = \frac{1}{4}$ and the product of zeroes $\alpha\beta = -1$. The quadratic polynomial is $x^2 - (\frac{1}{4})x + (-1) = x^2 - \frac{1}{4}x - 1$. To get integer coefficients, we can choose $k=4$. The polynomial becomes $4(x^2 - \frac{1}{4}x - 1) = 4x^2 - x - 4$.

(ii) Sum = $\sqrt{2}$, Product = $\frac{1}{3}$ Let the sum of zeroes $\alpha+\beta = \sqrt{2}$ and the product of zeroes $\alpha\beta = \frac{1}{3}$. The quadratic polynomial is $x^2 - (\sqrt{2})x + \frac{1}{3}$. To clear the fraction, we can choose $k=3$. The polynomial becomes $3(x^2 - \sqrt{2}x + \frac{1}{3}) = 3x^2 - 3\sqrt{2}x + 1$.

(iii) Sum = $0$, Product = $\sqrt{5}$ Let the sum of zeroes $\alpha+\beta = 0$ and the product of zeroes $\alpha\beta = \sqrt{5}$. The quadratic polynomial is $x^2 - (0)x + \sqrt{5} = x^2 + \sqrt{5}$.

(iv) Sum = $1$, Product = $1$ Let the sum of zeroes $\alpha+\beta = 1$ and the product of zeroes $\alpha\beta = 1$. The quadratic polynomial is $x^2 - (1)x + 1 = x^2 - x + 1$.

(v) Sum = $-\frac{1}{4}$, Product = $\frac{1}{4}$ Let the sum of zeroes $\alpha+\beta = -\frac{1}{4}$ and the product of zeroes $\alpha\beta = \frac{1}{4}$. The quadratic polynomial is $x^2 - (-\frac{1}{4})x + \frac{1}{4} = x^2 + \frac{1}{4}x + \frac{1}{4}$. To get integer coefficients, we can choose $k=4$. The polynomial becomes $4(x^2 + \frac{1}{4}x + \frac{1}{4}) = 4x^2 + x + 1$.

(vi) Sum = $4$, Product = $1$ Let the sum of zeroes $\alpha+\beta = 4$ and the product of zeroes $\alpha\beta = 1$. The quadratic polynomial is $x^2 - (4)x + 1 = x^2 - 4x + 1$.