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- Exercise 6.1 @matric:9th_science
- 6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)\\ &=(x-6)(x+2) \end{align}$ H.C.F= $x+2$ (ii) $\begin{align} x^3-27 &=x^3-3^3,\\ &=(x-3)(x^2+3x+9)\end{align}$ $\begin{align} x^2+6x-27&=x^2+9x-3x-27,\\ &=x(x+9)-3(x+9)\\ &=(x+6)(x-3) \end{align}$ $\begin{align} 2x^2-18&=2(x^2-9),\\ &=2(
- Exercise 2.6 (Solutions) @matric:9th_science:unit_02
- i) &= 2+3i+7-2i\\ &= 2+7+3i-2i\\ &= 9+i \end{array}$$ (ii) $$\begin{array}{cl} 2(5+4i)-3(7+4i... +8i-21-12i\\ &= 10-21+8i-12i\\ &= -11-4i \end{array}$$ (iii) $$\begin{array}{cl} -(-3+5i)-(4+9... = 3-5i-4-9i\\ &= 3-4-5i-9i\\ &= -1-14i \end{array}$$ (iv) $$\begin{array}{cl} 2i^2+6i^3+3i^{... imes12}(i)\\ &= -2 -6i+3+6i+4i\\ &= 1+4i \end{array}$$ ====Question 3==== * Simplify and writ
- Exercise 6.3 @matric:9th_science
- x-3y)-3y(2x-3y)\\&= (2x-3y)(2x-3y)\\&= (2x-3y)^2 \end{align}$\\ $\begin{align} \sqrt{4x^2-12xy +9y^2}&= \pm (2x-3y) \end{align}$ (ii) $x^2-1+\frac{1}{4x^2}, (x\neq 0)$\\... 1}{2x})+(\frac{1}{2x})^2\\&= (x-\frac{1}{2x})^2 \end{align}$\\ $\begin{align} \sqrt{x^2-1+\frac{1}{4x^2}}&= \pm (x-\frac{1}{2x}) \end{align}$ (iii) $\frac{1}{16}x^2-\frac{1}{12}xy+
- Exercise 2.5 (Solutions) @matric:9th_science:unit_02
- (i^2)^3\cdot i\\ &= {-1}^3 \cdot i\\ &= -i \end{array}$$ (ii) $$\begin{array}{cl} i^{50} &= (i^2 )^{25}\\ &= {-1}^{25}\\ &= -1 \end{array}$$ (iii) $$\begin{array}{cl} i^{12} &= (i^2 )^6\\ &= {-1}^6\\ &= 1 \end{array}$$ (iv) $$\begin{array}{cl} (-i)^8 &= (-i^2 )^4\\ &= {-1}^4\\ &= 1 \end{array}$$ (v) $$\begin{array}{cl} (-i)^5 &= (-i)^
- Review exercise @matric:9th_science
- 4)\\&=2 \times 2\times 2 \times (x^2+4)(x-2)(x+2)\end{align}$\\ $\begin{align}12 x^3-96&=12(x^3-8)\\&=2 \times 2\times 3 \times (x-2)(x^2+ 2 x+4)\end{align}$\\ $\begin{align} H.C.F. &= 2 \times 2 (x-2)\\&= 4(x-2)\end{align}$\\ ====Question 4:==== Find the $L.C.M... = 3(4x^2-25)\\&=3[(2x)^2-(5)^2]\\&= 3(2x+5)(2x-5)\end{align}$\\ $\begin{align}6x^2-13x-5 &= 6x^2-15x
- Exercise 6.2 @matric:9th_science
- x+6}{x+3}\\&=\frac{2x+8}{x+3}\\&=2\frac{x+4}{x+3}\end{align}$ ====Question 2:==== $\left[\frac{x+1}{x... eft(\frac{2+1}{x^4-1}\right)\\&=\frac{12x}{x^4-1}\end{align}$\\ ====Question 3:==== $\frac{1}{x^2-8x+... )(x-5)}\\&=\frac{2x-2x-6+6}{(x-1)(x-3)(x-5)}\\&=0\end{align}$ ====Question 4:==== $\frac{1}{x^2-8x+15... )(x-5)}\\&=\frac{2x-2x-6+6}{(x-1)(x-3)(x-5)}\\&=0\end{align}$ ====Question 5:==== $\frac{x+3}{2x^2+9x
- Exercise 2.4 (Solutions) @matric:9th_science:unit_02
- ^3.\sqrt[3]{3}}\\ &= \frac{7}{27.\sqrt[3]{3}} \end{array}\end{array}$$ (ii) $$\begin{array}{cl} \left(2x^5y^{-4}\right)\left(-8x^{-3}y^2\right) &=... &= -16x^2y^{-2}\\ &= \frac{-16x^2}{y^2} \end{array}$$ (iii) $$\begin{array}{cl} \left(\frac{x... -18} z^{-12}}\\ &=\frac{x^{18} z^{12}}{y^6} \end{array}$$ (iv) $$\begin{array}{cl} \frac{\left(81
- Exercise 4.1 @matric:9th_science
- y^(3-1)z^(5-2)}{30x^(3-2)}\\&=\frac{4 y^2 z^3}{x}\end{align}$\\ **Solution:**\\ (ii) $\frac{8 a(x+1)}{2... {2\times 4a(x+1)}{2(x+1)(x-1)}\\&= \frac{4a}{x-1}\end{align}$\\ **Solution:**\\ (iii) $\frac{(x+y)^2-4x... n{align}\frac{x^2+y^2+2xy-4xy}{x^2-2xy+y^2}\\&= 1\end{align}$\\ **Solution:**\\ (iv) $\frac{(x^3-y^3)(x... ^2-2xy+y^2)}{(x^3-y^3)}\\&= x^2-2x+y^2\\&=(x-y)^2\end{align}$\\ **Solution:**\\ (v) $\frac{(x+2)(x^2-1)
- Exercise 11.1 (Solutions) @matric:9th_science:unit11
- }}-{{130}^{\circ }}\\ & m\angle A={{50}^{\circ }}\end{align} Also we have $m\angle A=m\angle C$ (Oppos... B=180^\circ- 40^\circ \\ & m\angle DAB=140^\circ \end{align} Also \begin{align} & m\angle DAB+m\angle B... le B=180^\circ-140^\circ \\ & m\angle B=40^\circ \end{align} Now \begin{align} & m\angle B =m\angle D=... htarrow \,\,\,\,\,\,\,\,\, & m\angle D=40^\circ \end{align} Also \begin{align} & m\angle C=m\angle DA
- Exercise 2.3 (Solutions) @matric:9th_science:unit_02
- {-5^3}\\ & = {-5}^{3\times\frac{1}{3}}\\ &= {-5} \end{array}$$ (ii) $$\begin{array}{cl} \sqrt[4]{32}... {1}{4}\right)\times\sqrt[4]{2}\\ &= 2\sqrt[4]{2} \end{array}$$ (iii) $$\begin{array}{cl} \sqrt[5]{\f... }}\right)^\frac{1}{5}\\ &= \frac{\sqrt[5]{3}}{2} \end{array}$$ (iv) $$\beg... eft(\frac{-2}{3}\right)^{3\times\frac{1}{3}}\\ &= \frac{-2}{3} \end{array}$$
- Mathematics 9 (Science Group)
- & \text{if } a\geq 0, \\ -a & \text{if } a<0. \end{matrix}\right. $$ e.g. $|6|=6$, $|0|=0$, $|-3|=3... isector of a line segment is equidistant from its end points. * prove that any point equidistant from the end points of a line segment is on the right bisector
- Exercise 2.2 (Solutions) @matric:9th_science:unit_02
- . ... ... (iii)\\ &= 3y ... ... ... (iv) \end{array} $$ **Solution**\\ * (i) Distributive pr