$\begin{array}{l} 4 x^{2}-9 x=100 \\ \Rightarrow 4 x^{2}-9 x-100=0 \\ \Rightarrow 4 x^{2}-(25 x-16 x)-100=0 \\ \Rightarrow 4 x^{2}-25 x+16 x-100=0 \\ \Rightarrow x(4 x-25)+4(4 x-25)=0 \\...
Solution: The given system of eq. is: $\begin{array}{ll} 4 \mathrm{x}-3 \mathrm{y}=8 & \ldots \ldots \text { (i) } \\ 6 \mathrm{x}-\mathrm{y}=\frac{29}{3} & \ldots \ldots \text { (ii) }...
We write, $-2 x=-3 x+x$ as $3 x^{2} \times(-1)=-3 x^{2}=(-3 x) \times x$ $\begin{array}{l} \therefore 3 x^{2}-2 x-1=0 \\ \Rightarrow 3 x^{2}-3 x+x-1=0 \\ \Rightarrow 3 x(x-1)+1(x-1)=0 \\...
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
$\begin{array}{l} 6 x^{2}+x-12=0 \\ \Rightarrow 6 x^{2}+9 x-8 x-12=0 \\ \Rightarrow 3 x(2 x+3)-4(2 x+3)=0 \\ \Rightarrow(3 x-4)(2 x+3)=0 \\ \Rightarrow 3 x-4=0 \text { or } 2 x+3=0 \\ \Rightarrow...
$\begin{array}{l} 6 x^{2}+11 x+3=0 \\ \Rightarrow 6 x^{2}+9 x+2 x+3=0 \\ \Rightarrow 3 x(2 x+3)+1(2 x+3)=0 \\ \Rightarrow(3 x+1)(2 x+3)=0 \\ \Rightarrow 3 x+1=0 \text { or } 2 x+3=0 \\ \Rightarrow...
Solution: The given eq. are: $\begin{array}{l} \frac{x}{3}+\frac{y}{4}=11 \\ \Rightarrow 4 x+3 y=132 \ldots \ldots(i) \end{array}$ and $\frac{5 x}{6}-\frac{y}{3}+7=0$ $\Rightarrow 5 x-2 y=-42 \ldots...
using the relationship between the zeroes of he quadratic polynomial. => Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text...
$\begin{array}{l} x^{2}=18 x-77 \\ \Rightarrow x^{2}-18 x+77=0 \end{array}$ $\begin{array}{l} \Rightarrow x^{2}-(11 x+7 x)+77=0 \\ \Rightarrow x^{2}-11 x-7 x+77=0 \\ \Rightarrow x(x-11)-7(x-11)=0 \\...
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
Solution: The given system of eq. can be written as: $\begin{array}{cc} 9 \mathrm{x}-2 \mathrm{y}=108 & \ldots \ldots \text { (i) } \\ 3 \mathrm{x}+7 \mathrm{y}=105 & \ldots \text {..(ii) }...
We write, $-3 x=3 x-6 x$ as $9 x^{2} \times(-2)=-18 x^{2}=3 x \times(-6 x)$ $\begin{array}{l} \therefore 9 x^{2}-3 x-2=0 \\ \Rightarrow 9 x^{2}+3 x-6 x-2=0 \\ \Rightarrow 3 x(3 x+1)-2(3 x+1)=0 \\...
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
We write, $6 x=x+5 x$ as $x^{2} \times 5=5 x^{2}=x \times 5 x$ $\begin{array}{l} \therefore x^{2}+6 x+5=0 \\ \Rightarrow x^{2}+x-5 x+5=0 \end{array}$ $\begin{array}{l} \Rightarrow x(x+1)+5(x+1)=0 \\...
For finding the zeroes of the quadratic polynomial we will equate $f(x)$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}$ are $-\frac{2}{\sqrt{3}}$ or...
Solution: The given system of eq. is: $\begin{array}{l} 2 x-y+3=0 \ldots \ldots(i) \\ 3 x-7 y+10=0 \ldots \ldots(i i) \end{array}$ From equation(i), write y in terms of $x$ to obtain $y=2 x+3$ On...
We write, $x=4 x-3 x$ as $2 x^{2} \times(-6)=-12 x^{2}=4 x \times(-3 x)$ $\begin{array}{l} \therefore 2 x^{2}+x-6=0 \\ \Rightarrow 2 x^{2}+4 x-3 x-6=0 \\ \Rightarrow 2 x(x+2)-3(x+2)=0 \\...
$\begin{array}{l} 3 x^{2}-243=0 \\ \Rightarrow 3\left(x^{2}-81\right)=0 \\ \Rightarrow(x)^{2}-(9)^{2}=0 \\ \Rightarrow(x+9)(x-9)=0 \\ \Rightarrow x+9=0 \text { or } x-9=0 \\ \Rightarrow x=-9 \text {...
To find the zeroes of the quadratic polynomial we will equate $\mathrm{f}(\mathrm{x})$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=6 x^{2}-3$ are...
Solution: The given system of equation is: $\begin{array}{c} 3 \mathrm{x}-5 \mathrm{y}-19=0 \\ -7 \mathrm{x}+3 \mathrm{y}+1=0 \end{array}$ When multiplying equation(i) by 3 and equation(ii) by 5 ,...
$\begin{array}{l} (2 x-3)(3 x+1)=0 \\ \Rightarrow 2 x-3=0 \text { or } 3 x+1=0 \\ \Rightarrow 2 x=3 \text { or } 3 x=-1 \\ \Rightarrow x=\frac{3}{2} \text { or } x=-\frac{1}{3} \end{array}$ As a...
Solution: The given system of eq. is: $\begin{array}{cc} 2 \mathrm{x}-3 \mathrm{y}=13 & \ldots \ldots \text { (i) } \\ 7 \mathrm{x}-2 \mathrm{y}=20 & \ldots \ldots \text { (ii) }...
We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula $\mathrm{x}^{2}-($ sum of the zeroes $) \mathrm{x}+$ product of zeroes $\Rightarrow...
"If $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are two polynomials such that degree of $\mathrm{f}(\mathrm{x})$ is greater than degree of $\mathrm{g}(\mathrm{x})$ where...
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
Solution: The given system of equation is: $\begin{array}{l} 2 x+3 y=0 & \ldots \ldots \text { (i) } \\ 3 x+4 y=5 & \ldots \text {..(ii) } \end{array}$ When multiplying equation(i) by 4 and...
Equating $\mathrm{x}^{2}-\mathrm{x}$ to 0 to find the zeroes, we will get x(x-1)=0 x(x-1)=0 ⇒x=0 or x-1=0 \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}-1=0...
using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{X}=-\mathrm{a}$ Since, $(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ Hence, It will satisfy the above polynomial...
Solution: The given system of equations is $x-y=3 \dots \dots(i)$ $\frac{x}{3}+\frac{y}{2}=6 \dots \dots(ii)$ From equation(i), write y in terms of $x$ to obtain $y=x-3$ When substituting...
using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...
$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...
$x=-2$ is one zero of the polynomial $3 x^{2}+4 x+2 k$ Therefore, it will satisfy the above polynomial. Now, we have $3(-2)^{2}+4(-2) 1+2 k=0$ $\Rightarrow 12-8+2 k=0$ $\Rightarrow...
$x=1$ is one zero of the polynomial $a x^{2}-3(a-1) x-1$ Therefore, it will satisfy the above polynomial. Now, we have $a(1)^{2}-(a-1) 1-1=0$ $\Rightarrow a-3 a+3-1=0$ $\Rightarrow-2 \mathrm{a}=-2$...
Solution: The given system of equation is: $\mathrm{x}+\mathrm{y}=3 \ldots \ldots \text { (i) }$ $4 x-3 y=26 \ldots \ldots \text { (ii) }$ When multiplying equation(i) by 3 , we obtain: $3 x+3 y=9...
$x=-4$ is one zero of the polynomial $x^{2}-x-(2 k+2)$ Therefore, it will satisfy the above polynomial. Now, we have $(-4)^{2}-(-4)-(2 k+2)=0$ $\Rightarrow 16+4-2 \mathrm{k}-2=0$ $\Rightarrow 2...
$x=3$ is one zero of the polynomial $2 x^{2}+x+k$ Therefore, it will satisfy the above polynomial. Now, we have $2(3)^{2}+3+k=0$ $\Rightarrow 21+\mathrm{k}=0$ $\Rightarrow...
$x=2$ is one zero of the quadratic polynomial $k x^{2}+3 x+k$ Therefore, it will satisfy the above polynomial. $k(2)^{2}+3(2)+k=0$ $\Rightarrow 4 \mathrm{k}+6+\mathrm{k}=0$ $\Rightarrow 5...
If the zeroes of the quadratic polynomial are $\alpha$ and $\beta$ then the quadratic polynomial can be found as $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...
Solution: From the first eq., write $y$ in terms of $x$ $y=2-2 x\dots \dots(i)$ Substituting different values of $\mathrm{x}$ in equation(i) to get different values of $\mathrm{y}$ For $x=0,...
$f(x)=x^{2}-3 x-m(m+3)$ adding and subtracting $\mathrm{mx}$, $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$...
On subtracting $x=\left(-\frac{1}{2}\right)$ in the given equation, we get $\begin{array}{l} \text { L.H.S. }=3 x^{2}+2 x-1 \\ =3 \times\left(-\frac{1}{2}\right)^{2}+2...
$f(x)=x^{2}+x-p(p+1)$ adding and subtracting $\mathrm{px}$, we get $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$...
Let the other zeroes of $x^{2}-4 x+1$ be a (using the relationship between the zeroes of the quadratic polynomial) sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coef ficient of }...
Solution: From the first equation, write y in terms of $x$ $y=6-2 x\dots \dots(i)$ Substituting different values of $x$ in equation(i) to get different values of $y$ For $x=0, y=6-0=6$ For $x=2,...
$f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. Since $(3+\sqrt{2})$ and $(3-\sqrt{2})$ are the zeroes of $f(x)$ it follows that each one of $(x+3+\sqrt{2})$ and $(x+3-\sqrt{2})$ is a factor of $f(x)$...
(i) $\begin{array}{l} (x+2)^{3}=x^{3}-8 \\ \Rightarrow x^{3}+6 x^{2}+12 x+8=x^{3}-8 \\ \Rightarrow 6 x^{2}+12 x+16=0 \end{array}$ This is of the form $a x^{2}+b x+c=0$ Hence, the given equation is a...
The given polynomial is $f(x)=x^{4}+4 x^{3}-2 x^{2}-20 x-15$. Since $(x-\sqrt{5})$ and $(x+\sqrt{5})$ are the zeroes of $f(x)$ it follows that each one of $(x-\sqrt{5})$ and $(x$ $+\sqrt{5})$ is a...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the x-axis and y-axis, respectively. Graph of $2 x+3 y=4$ $\begin{array}{l} 2 x+3 y=4 \\ \Rightarrow 3 y=(-2...
The given polynomial is $f(x)=2 x^{4}-3 x^{3}-5 x^{2}+9 x-3$ Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and $(x+\sqrt{3})$ is a factor of...
(i) Clearly, $\left(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\right)$ is a quadratic polynomial. $\therefore \sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ is a quadratic equation. (ii) Clearly, $\left(\frac{1}{3}...
Solution: From the first eq., write y in terms of $x$ $\mathrm{y}=\frac{x-6}{2}\dots \dots(i)$ Substituting different values of $x$ in equation(i) to get different values of y For $x=-2,...
(i) $\left(x^{2}-x+3\right)$ is a quadratic polynomial $\therefore x^{2}-x+3=0$ is a quadratic equation. (ii) Clearly, $\left(2 x^{2}+\frac{5}{2} x-\sqrt{3}\right)$ is a quadratic polynomial....
Solution: From the first eq., write y in terms of $x$ $y=\frac{x-5}{2}\dots \dots(i)$ Substituting different values of $x$ in eq.(i) to get different values of y For $x=-5, y=\frac{-5-5}{2}=-5$ For...
Solution: Draw a horizontal line on a graph paper $\mathrm{X}^{\prime} \mathrm{OX}$ and a vertical line YOY' representing the $\mathrm{x}-$ axis and $y$-axis, respectively. Graph of $2...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' representing the xaxis and y-axis, respectively. Graph of $3 x-y=5$ $3 x-y=5$ $\Rightarrow y=3 x-5\dots \dots(i)$...
Solution: From the first eq., write y in terms of $x$ $y=\frac{6-2 x}{3}\dots \dots(i)$ Substituting different values of $x$ in equation(i) to get different values of y For $x=-3, y=\frac{6+6}{3}=4$...
Let f(x)=x4+x3-23x2-3x+60 \text { Let } f(x)=x^{4}+x^{3}-23 x^{2}-3 x+60 Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and...
Let $f(x)=x^{4}+x^{3}-34 x^{2}-4 x+120$ Since 2 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-2)$ and $(x+2)$ is a factor of $f(x)$ Consequently,...
Since 3 and $-3$ are the zeroes of $f(x)$, it follows that each one of $(x+3)$ and $(x-3)$ is a factor of $f(x)$ Consequently, $(x-3)(x+3)=\left(x^{2}-9\right)$ is a factor of $f(x)$. On dividing...
Let $f(x)=x^{3}-4 x^{2}-7 x+10$ Since 1 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-1)$ and $(x+2)$ is a factor of $f(x)$ Consequently, $(x-1)(x+2)=\left(x^{2}+x-2\right)$ is...
Let $f(x)=x^{3}+2 x^{2}-11 x-12$ Since $-1$ is a zero of $f(x),(x+1)$ is a factor of $f(x)$. On dividing $\mathrm{f}(\mathrm{x})$ by $(\mathrm{x}+1)$, we get $$ \begin{aligned} &f(x)=x^{3}+2...
using division rule, Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore 3 x^{3}+x^{2}+2 x+5=(3 x-5) g(x)+9 x+10$ $\Rightarrow 3 x^{3}+x^{2}+2 x+5-9 x-10=(3 x-5) g(x)$ $\Rightarrow 3...
Let $f(x)=2 x^{4}+3 x^{3}-2 x^{2}-9 x-12$ and $g(x)$ as $x^{2}-3$
Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. $\text { Graph of } 4 x-3 y+4=0$ $\begin{array}{l} 4 x-3 y+4=0...
Solution: From the first eq., write y in terms of $x$ $y=\frac{2 x-12}{3}\dots \dots (i)$ Substituting different values of $\mathrm{x}$ in eq.(i) to get different values of $\mathrm{y}$ For $x=0,...
Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. $\begin{array}{l} \quad \text { Graph of } 2 x-5 y+4=0 \\ 2...
Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. $\text { Graph of } 2 x-y=1$ $\begin{array}{l} 2 x-y=1 \\...
Solution: Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ as the $x$-axis and $y$-axis, respectively. Graph of $4 x-y=4$ $\begin{array}{l} 4 x-y=4 \\...
Solution: Draw a horizontal line on a gfraph paper $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ as the $x$-axis and $y$-axis, respectively. Graph of $2 x-3 y-17=0$ $\begin{array}{l} 2 x-3...
Solution: On a graph paper, draw a horizontal line $X^{\prime} O X$ and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. Graph of $4 x-5 y+16=0$ $\begin{array}{l} 4 x-5 y+16=0 \\...
Solution: Draw a horizontal line on a graph paper $\mathrm{X}^{\prime} \mathrm{OX}$ and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. Graph of...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. Graph of $4 x-3 y+4=0$ $\begin{array}{l} 4 x-3 y+4=0 \\...
Solution: From the first eq., write y in terms of $x$ $\mathrm{y}=\frac{2 x+4}{3}\dots \dots(i)$ Substituting different values of $x$ in eq.(i) to get different values of $y$ For $x=-2,...
Solution: From the first eq., write y in terms of $\mathrm{x}$ $y=x+3\dots \dots(i)$ Substituting different values of $x$ in eq(i) to get different values of $y$ For $x=-3, y=-3+3=0$ For $x=-1,...
$f(x)$ as $x^{4}+0 x^{3}+0 x^{2}-5 x+6$ and $g(x) a s-x^{2}+2$ Quotient $q(x)=-x^{2}-2$ Remainder $\mathrm{r}(\mathrm{x})=-5 \mathrm{x}+10$
(i) Which firm A or B pays out the larger amount as weekly wages? (ii) Which firm A or B has greater variability in individual wages? Solution:
Quotient $q(x)=x^{2}+x-3$ Remainder $r(x)=8$
Quotient $q(x)=x-\overline{3}$ Remainder $r(x)=7 x-9$ 7. If $f(x)=x^{4}-3 x^{2}+4 x+5$ is divided by $g(x)=x^{2}-x+1$
sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as $x^{3}-($ sum of the zeroes $) x^{2}+($ sum of the...
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=\frac{1}{2}, b=1$ and $c=-3$ Substituting the values...
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=2, b=-3$ and $c=4$ Substituting the values in 1 , we...
p(x)=3x3-10x2-27x+10 p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right) p(5)=3×53-10×52-27×5+10=(375-250-135+10)=0 p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times...
The given polynomial is $p(x)=\left(x^{3}-2 x^{2}-5 x+6\right)$ $$ \begin{aligned} &\therefore \mathrm{p}(3)=\left(3^{3}-2 \times 3^{2}-5 \times 3+6\right)=(27-18-15+6)=0 \\...
$x=\frac{2}{3}$ is one of the zero of $3 x^{3}+16 x^{2}+15 x-18$ Now, we have $\mathrm{x}=\frac{2}{3}$ $\Rightarrow \mathrm{x}-\frac{2}{3}=0$ Now, we divide $3 x^{3}+16 x^{2}+15 x-18$ by...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. $\begin{array}{l} x+2 y+2=0 \\ \Rightarrow 2 y=(-2-x) \\...
Solution: Draw a horizontal line oon a graph paper X'OX and a vertical line YOY' as the $x$-axis and $y$-axis, respectively. Graph of $2 x+3 y=4$ $2 x+3 y=4$ $\Rightarrow 3 y=(4-2 x)$ $\therefore...
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{x}=-\mathrm{a}$ Since, it satisfies the above polynomial. => $2(-a)^{2}+2 a(-a)+5(-a)+10=0$ $\Rightarrow 2 a^{2}-2...
$a x^{2}+7 x+b=0$ Since, $x=\frac{2}{3}$ is the root of the above quadratic equation Hence, it will satisfy the above equation. => $a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$...
Solution: From the first eq., write y in terms of $x$ $y=\frac{2 x+13}{3}\dots \dots (i)$ Substituting different values of $x$ in eq.(i) to get different values of $y$ For $x=-5,...
Quadratic equation can be found if we know the sum of the roots and product of the roots by using the formula: $\mathrm{x}^{2}-($ Sum of the roots) $\mathrm{x}+$ Product of roots $=0$ $\Rightarrow...
Solution: From the first eq., write y in terms of $x$ $y=-\left(\frac{5+2 x}{3}\right)\dots \dots (i)$ Substitute different values of $\mathrm{x}$ in eq.(i) to get different values of $\mathrm{y}$...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. =>$(\alpha+\beta)=\frac{5}{2}$ and $\alpha \beta=1$ $\therefore...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and y-axis, respectively. Graph of $3 \mathbf{x}+\mathbf{y}+\mathbf{1}=\mathbf{0}$...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $f(x)$. Then $(\alpha+\beta)=0$ and $\alpha \beta=-1$ $\therefore f(x)=x^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\Rightarrow...
Solution: The given eq. are: $3 x+2 y=12 \dots\dots (i)$ $5 x-2 y=4 \dots \dots(ii)$ From eq.(i), write y in terms of $x$ $\mathrm{y}=\frac{12-3 x}{2}\dots \dots(iii)$ Substitute different values of...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and y-axis, respectively. $\begin{array}{l} 2 x-5 y+4=0 \\ \Rightarrow 5 y=(2 x+4) \\...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' as the $\mathrm{x}$-axis and $\mathrm{y}$-axis, respectively. $\begin{array}{l} 2 x+3 y=8 \\ \Rightarrow 3 y=(8-2 x)...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. Then $(\alpha+\beta)=8$ and $\alpha \beta=12$ $\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$...
Let $\alpha=\frac{2}{3}$ and $\beta=\frac{-1}{4}$. Sum of the zeroes $=(\alpha+\beta)=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}$ Product of the zeroes, $\alpha...
Solution: Draw a horizontal line on a graph paper X'OX and a vertical line YOY' representing the xaxis and $y$-axis, respectively. Graph of $3 x+2 y=4$ $3 x+2 y=4$ $\Rightarrow 2 y=(4-3 x)$...
Solution: Draw a horizontal line on a graph paper$\mathrm{X}^{\prime} \mathrm{OX}$ and a vertical line YOY' representing the $\mathrm{x}-$ axis and $\mathrm{y}$-axis, respectively. $\begin{array}{l}...
$$ \begin{aligned} &3 x^{2}-x-4=0 \\ &\Rightarrow 3 x^{2}-4 x+3 x-4=0 \\ &\Rightarrow x(3 x-4)+1(3 x-4)=0 \\ &\Rightarrow(3 x-4)(x+1)=0 \\ &\Rightarrow(3 x-4) \text { or }(x+1)=0 \\ &\Rightarrow...
f(u)=5u2+10u \mathrm{f}(\mathrm{u})=5 \mathrm{u}^{2}+10 \mathrm{u} It can be written as $5 \mathrm{u}(\mathrm{u}+2)$ ∴f(u)=0⇒5u=0 or u+2=0 \therefore \mathrm{f}(\mathrm{u})=0...
$$ \begin{aligned} &\mathrm{f}(\mathrm{x})=8 \mathrm{x}^{2}-4 \\ &\text { It can be written as } 8 \mathrm{x}^{2}+0 \mathrm{x}-4 \\ &=4\left\{(\sqrt{2} x)^{2}-(1)^{2}\right\} \\ &=4(\sqrt{2}...
$$ \begin{aligned} &4 x^{2}-4 x+1=0 \\ &\Rightarrow(2 x)^{2}-2(2 x)(1)+(1)^{2}=0 \end{aligned} $$ $$ \begin{aligned} &\Rightarrow(2 \mathrm{x}-1)^{2}=0 \quad\left[\because \mathrm{a}^{2}-2...
Solution: By using the formula for standard deviation: $\begin{array}{l} \mathrm{SD}=\sqrt{\operatorname{Var}(\mathrm{X})} \\ \text { Mean }=\sum \frac{\mathrm{f}_{\mathrm{i}}...
$$ \begin{aligned} \mathrm{f}(\mathrm{x}) &=2 \mathrm{x}^{2}-11 \mathrm{x}+15 \\ &=2 \mathrm{x}^{2}-(6 \mathrm{x}+5 \mathrm{x})+15 \\ &=2 \mathrm{x}^{2}-6 \mathrm{x}-5 \mathrm{x}+15 \\ =& 2...
$$ \begin{aligned} &2 \sqrt{3} x^{2}-5 x+\sqrt{3} \\ &\Rightarrow 2 \sqrt{3} x^{2}-2 x-3 x+\sqrt{3} \\ &\Rightarrow 2 x(\sqrt{3} x-1)-\sqrt{3}(\sqrt{3} x-1)=0 \\ &\Rightarrow(\sqrt{3} x-1) \text {...
$x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow \mathrm{x}=-4$ or $\mathrm{x}=-3$ Sum of...
Solution: Let's suppose that there exist a smallest positive integer which is neither odd nor even, say $n .$ As $n$ is least positive integer which is neither even nor odd, $n-1$ must be either odd...
Solution: (i) When applying Euclid's algorithm, that is dividing $2520$ by $405$, we obtain, Quotient $=6$, Remainder $=90$ $\therefore 2520=405 \times 6+90$ Again upon applying Euclid's algorithm,...
Solution: It is given that, Dividend $=1365$, Quotient $=31$, Remainder Let the divisor be $x$ $\begin{array}{l}\text { Dividend }=\text { Divisor } \times \text { Quotient }+\text { Remainder } \\...
Solution: It is known to us that, Dividend $=$ Divisor $\times$ Quotient $+$ Remainder It is given that, Divisor $=61$, Quotient $=27$, Remainder $=32$ Let the Dividend be $x$ $\begin{aligned}...
Answers: (i) Using prime factorization, 96 = 25 × 3 404 = 22 × 101 HCF = 22 HCF = 4 LCM = 25 × 3 × 101 LCM = 9696 (ii) Using prime factorization, 144 = 24 × 32 198 = 2 × 32 × 11 HCF = 2 × 32 HCF =...
The equation passes through (2, 3) and make an angle of 450with the line 3x + y – 5 = 0. Since, the equations of two lines passing through a point x1,y1 and making an angle α with the given line y =...
Solution: The area required is represented by the shaded area OBAO as The point of intersection of the curves $y=x$ and $y=x^{2}$ is A(1,1) Draw AC perpendicular to x-axis $\therefore$ Area...
The correct option is (iii) 8×104 Nm–2
Solution: Given that, \[sin\text{ }A\text{ }=\text{ }9/41~\ldots \ldots \ldots \ldots .\text{ }\left( 1 \right)\] Required to find: \[cos\text{ }A,\text{ }tan\text{ }A\] By definition, we...
Let ‘a’ be any positive integer. Then from Euclid’s division lemma, a = bq+r; where 0 < r < b Putting b=6 we get, ⇒ a = 6q + r, 0 < r < 6 For r = 0, we get\[a=6q=2\left( 3q...
\[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\] \[\mathbf{2x}\text{ }\text{ }\mathbf{3y}\text{ }\text{ }\mathbf{12}\text{ }=\text{...
(i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3)(iii) (a, b), (- a, – b) We know that formula to find the distance (d) between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}}...