Solving the following quadratic equations by factorization method:
Question: Solving the following quadratic equations by factorization method: (i) $x^{2}+10 i x-21=0$ (ii) $x^{2}+(1-2 i) x-2 i=0$ (iii) $x^{2}-(2 \sqrt{3}+3 i) x+6 \sqrt{3} i=0$ (iv) $6 x^{2}-17 i x-12=0$ Solution: (i) $x^{2}+10 i x-21=0$ $\Rightarrow x^{2}+7 i x+3 i x-21=0$ $\Rightarrow x(x+7 i)+3 i(x+7 i)=0$ $\Rightarrow(x+7 i)(x+3 i)=0$ $\Rightarrow(x+7 i)=0$ or $(x+3 i)=0$ $\Rightarrow x=-7 i,-3 i$ So, the roots of the given quadratic equation are $-3 i$ and $-7 i$. (ii) $x^{2}+(1-2 i) x-2 i...
Read More →Factorize:
Question: Factorize: $2 x+4 y-8 x y-1$ Solution: We have: $2 x+4 y-8 x y-1=(2 x-8 x y)-(1-4 y)$ $=2 x(1-4 y)-1(1-4 y)$ $=(1-4 y)(2 x-1)$...
Read More →Let R+ be the set of all non-negative real numbers.
Question: Let $R^{+}$be the set of all non-negative real numbers. If $f: R^{+} \rightarrow R^{+}$and $g: R^{+} \rightarrow R^{+}$are defined as $f(x)=x^{2}$ and $g(x)=+\sqrt{x}$, find fog and gof. Are they equal functions? Solution: Given, $f: R^{+} \rightarrow R^{+}$and $g: R^{+} \rightarrow R^{+}$ So, fog: $R^{+} \rightarrow R^{+}$and gof: $R^{+} \rightarrow R^{+}$ Domains offogandgofare the same $(f o g)(x)=f(g(x))=f(\sqrt{x})=(\sqrt{x})^{2}=x$ $(g o f)(x)=g(f(x))=g\left(x^{2}\right)=\sqrt{x^...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $4 x^{2}+4 \sqrt{3} x+3=0$ Solution: We have been given that, $4 x^{2}+4 \sqrt{3} x+3=0$ Now divide throughout by 4. We get, $x^{2}+\sqrt{3} x+\frac{3}{4}=0$ Now take the constant term to the RHS and we get $x^{2}+\sqrt{3} x=-\frac{3}{4}$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}+2\left(\frac{\sqrt{3}}{2}\right) x+\left(\frac{\sqrt{3}}{2}\righ...
Read More →Factorize:
Question: Factorize: $x^{3}-x^{2}+a x+x-a-1$ Solution: We have: $x^{3}-x^{2}+a x+x-a-1=\left(x^{3}-x^{2}\right)+(a x-a)+(x-1)$ $=x^{2}(x-1)+a(x-1)+1(x-1)$ $=(x-1)\left(x^{2}+a+1\right)$...
Read More →Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $2 x^{2}+x+4=0$ Solution: We have been given that, $2 x^{2}+x+4=0$ Now divide throughout by 2. We get, $x^{2}+\frac{1}{2} x+2=0$ Now take the constant term to the RHS and we get $x^{2}+\frac{1}{2} x=-2$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}+\frac{1}{2} x+\left(\frac{1}{4}\right)^{2}=\left(\frac{1}{4}\right)^{2}-2$ $x^{2}+\left(\frac{1}{4}\...
Read More →Factorize:
Question: Factorize: $a b x^{2}+a^{2} x+b^{2} x+a b$ Solution: We have: $a b x^{2}+a^{2} x+b^{2} x+a b=\left(a b x^{2}+b^{2} x\right)+\left(a^{2} x+a b\right)$ $=b x(a x+b)+a(a x+b)$ $=(a x+b)(b x+a)$...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $2 x^{2}+x-4=0$ Solution: We have been given that, $2 x^{2}+x-4=0$ Now divide throughout by 2. We get, $x^{2}+\frac{1}{2} x-2=0$ Now take the constant term to the RHS and we get $x^{2}+\frac{1}{2} x=2$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}+\frac{1}{2} x+\left(\frac{1}{4}\right)^{2}=\left(\frac{1}{4}\right)^{2}+2$ $x^{2}+\left(\frac{1}{4}\r...
Read More →Factorize:
Question: Factorize: $3 a x-6 a y-8 b y+4 b x$ Solution: We have: $3 a x-6 a y-8 b y+4 b x=(3 a x-6 a y)+(4 b x-8 b y)$ $=3 a(x-2 y)+4 b(x-2 y)$ $=(x-2 y)(3 a+4 b)$...
Read More →Show that $(p-1)$ is a factor of $\left(p^{10}-1\right)$ and also of $\left(p^{11}-1\right)$.
[question] Question. Show that $(p-1)$ is a factor of $\left(p^{10}-1\right)$ and also of $\left(p^{11}-1\right)$. [/question] [solution] Solution: Let $f(p)=p^{10}-1$ and $g(p)=p^{11}-1$ Putting $p=1$ in $f(p)$, we get $f(1)=1^{10}-1=1-1=0$ Therefore, by factor theorem, $(p-1)$ is a factor of $\left(p^{10}-1\right)$. Now, putting $p=1$ in $g(p)$, we get $g(1)=1^{11}-1=1-1=0$ Therefore, by factor theorem, $(p-1)$ is a factor of $\left(p^{11}-1\right)$. [/solution]...
Read More →Find fog (2) and gof (1) when
Question: Find $f o g(2)$ and $g o f(1)$ when : $f: R \rightarrow R ; f(x)=x^{2}+8$ and $g: R \rightarrow R ; g(x)=3 x^{3}+1$ Solution: $(f o g)(2)=f(g(2))=f\left(3 \times 2^{3}+1\right)=f(25)=25^{2}+8=633$ $(g o f)(1)=g(f(1))=g\left(1^{2}+8\right)=g(9)=3 \times 9^{3}+1=2188$...
Read More →Factorize:
Question: Factorize: $a^{3}+a-3 a^{2}-3$ Solution: We have: $a^{3}+a-3 a^{2}-3=\left(a^{3}-3 a^{2}\right)+(a-3)$ $=a^{2}(a-3)+1(a-3)$ $=(a-3)\left(a^{2}+1\right)$...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $3 x^{2}+11 x+10=0$ Solution: We have been given that, $3 x^{2}+11 x+10=0$ Now divide throughout by 3. We get, $x^{2}+\frac{11}{3} x+\frac{10}{3}=0$ Now take the constant term to the RHS and we get $x^{2}+\frac{11}{3} x=-\frac{10}{3}$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}+\frac{11}{3} x+\left(\frac{11}{6}\right)^{2}=\left(\frac{11}{6}\righ...
Read More →Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $2 x^{2}-7 x+3=0$ Solution: We have to find the roots of given quadratic equation by the method of completing the square. We have, $2 x^{2}-7 x+3=0$ We should make the coefficient of $x^{2}$ unity. So, $x^{2}-\frac{7}{2} x+\frac{3}{2}=0$ Now shift the constant to the right hand side, $x^{2}-\frac{7}{2} x=-\frac{3}{2}$ Now add square of half of coefficient ofon both the sides, $x^...
Read More →Factorize:
Question: Factorize: $(2 x-3)^{2}-8 x+12$ Solution: We have: $(2 x-3)^{2}-8 x+12=(2 x-3)^{2}-4(2 x-3)$ $=(2 x-3)[(2 x-3)-4]$ $=(2 x-3)(2 x-3-4)$ $=(2 x-3)(2 x-7)$...
Read More →Factorize:
Question: Factorize: $(3 a-1)^{2}-6 a+2$ Solution: We have: $(3 a-1)^{2}-6 a+2=(3 a-1)^{2}-2(3 a-1)$ $=(3 a-1)[(3 a-1)-2]$ $=(3 a-1)(3 a-1-2)$ $=(3 a-1)(3 a-3)$ $=3(3 a-1)(a-1)$...
Read More →Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $x^{2}-4 \sqrt{2 x}+6=0$ Solution: We have been given that, $x^{2}-4 \sqrt{2} x+6=0$ Now we take the constant term to the right hand side and we get $x^{2}-4 \sqrt{2} x=-6$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}-4 \sqrt{2} x+(2 \sqrt{2})^{2}=(2 \sqrt{2})^{2}-6$ $x^{2}+(2 \sqrt{2})^{2}-2(2 \sqrt{2}) x=2$ $(x-2 \sqrt{2})^{2}=2$ Since right ha...
Read More →Write the argument of
Question: Write the argument of $(1+i \sqrt{3})(1+i)(\cos \theta+i \sin \theta)$. Disclaimer: There is a misprinting in the question. It should be $(1+i \sqrt{3})$ instead of $(1+\sqrt{3})$. Solution: Let the argument of $(1+i \sqrt{3})$ be $\alpha$. Then, $\tan \alpha=\frac{\sqrt{3}}{1}=\tan \frac{\pi}{3}$ $\Rightarrow \alpha=\frac{\pi}{3}$ Let the argument of $(1+i)$ be $\beta$. Then, $\tan \beta=\frac{1}{1}=\tan \frac{\pi}{4}$ $\Rightarrow \beta=\frac{\pi}{4}$ Let the argument of $(\cos \thet...
Read More →Factorize:
Question: Factorize: $x^{2}+y-x y-x$ Solution: We have: $x^{2}+y-x y-x=\left(x^{2}-x y\right)-(x-y)$ $=x(x-y)-1(x-y)$ $=(x-y)(x-1)$...
Read More →Let A = {a, b, c}, B = {u v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as :
Question: LetA= {a,b,c},B= {uv,w} and letfandgbe two functions fromAtoBand fromBtoA,respectively, defined as : f= {(a,v), (b,u), (c,w)},g= {(u,b), (v,a), (w,c)}. Show thatfandgboth are bijections and findfogandgof. Solution: Provingfis a bijection: $f=\{(a, v),(b, u),(c, w)\}$ and $f: A \rightarrow B$ Injectivity off: No two elements ofAhave the same image in B.So,fis one-one.Surjectivity off: Co-domain off= {uv,w}Range off= {uv,w}Both are same.So,fis onto.Hence,fis a bijection. Provinggis a bij...
Read More →Factorize:
Question: Factorize: $p x-5 q+p q-5 x$ Solution: We have: $p x-5 q+p q-5 x=(p x-5 x)+(p q-5 q)$ $=x(p-5)+q(p-5)$ $=(p-5)(x+q)$...
Read More →Solve the following
Question: If $|z|=2$ and $\arg (z)=\frac{\pi}{4}$, find $z$ Solution: We know that, $z=|z|\{\cos [\arg (z)]+i \sin [\arg (z)]\}$ $=2\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ $=2\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)$ $=\sqrt{2}(1+i)$ Hence, $z=\sqrt{2}(1+i)$...
Read More →Solve the following
Question: If $|z|=2$ and $\arg (z)=\frac{\pi}{4}$, find $z$ Solution: We know that, $z=|z|\{\cos [\arg (z)]+i \sin [\arg (z)]\}$ $=2\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ $=2\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)$ $=\sqrt{2}(1+i)$ Hence, $z=\sqrt{2}(1+i)$...
Read More →Find the real value of a for which
Question: Find the real value of $a$ for which $3 i^{3}-2 a i^{2}+(1-a) i+5$ is real. Solution: $3 i^{3}-2 a i^{2}+(1-a) i+5$ $=-3 i+2 a+(1-a) i+5$ $=(2 a+5)+i(1-a-3)$ $=(2 a+5)+i(-2-a)$ Since, $3 i^{3}-2 a i^{2}+(1-a) i+5$ is real. $\therefore \operatorname{Im}\left[3 i^{3}-2 a i^{2}+(1-a) i+5\right]=0$ $\Rightarrow-2-a=0$ $\Rightarrow a=-2$ Hence, the real value of $a$ for which $3 i^{3}-2 a i^{2}+(1-a) i+5$ is real is $-2$....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq 4,7$ Solution: We have been given $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$ $\frac{-11}{x^{2}-3 x-28}=\frac{11}{30}$ $-30=x^{2}-3 x-28$ $x^{2}-3 x+2=0$ $x^{2}-2 x-x+2=0$ $x(x-2)-1(x-2)=0$ $(x-1)(x-2)=0$ Therefore, $x-1=0$ $x=1$ or, $x-2=0$ $x=2$ Hence, $x=1$ or $x=2$....
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