If $u_{i}=\frac{x_{i}-25}{10}, \Sigma f_{i} u_{i}=20, \Sigma f_{i}=100$, then $\bar{x}=$
[question] Question. If $u_{i}=\frac{x_{i}-25}{10}, \Sigma f_{i} u_{i}=20, \Sigma f_{i}=100$, then $\bar{x}=$ (a) 23 (b) 24 (c) 27 (d) 25 [/question] [solution] solution: Given: $u_{i}=\frac{x_{i}-25}{10}, \Sigma f_{i} u_{i}=20$ and $\Sigma f_{i}=100$ Now, $u_{i}=\frac{x_{i}-A}{h}=\frac{x_{i}-25}{10}$ Therefore, $h=10$ and $A=25$ We know that $\bar{x}=A+h\left\{\frac{1}{N} \sum f_{i} u_{i}\right\}$ $=25+10\left\{\frac{1}{100} \times 20\right\}$ $=25+10 \times \frac{1}{5}$ $=25+2$ $\bar{x}=27$ He...
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Question: $2 x^{2}+x+1=0$ Solution: Given: $2 x^{2}+x+1=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=2, b=1$ and $c=1$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-1+\sqrt{1-4 \times 2 \times 1}}{2 \times 2} \quad$ and $\quad \beta=\frac{-1-\sqrt{1-4 \times 2 \times 1}}{2 \times 2}$ $\Rightarrow \alpha=\frac{-1+\sqrt{-7}}{4} \quad$ and ...
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Question: $13 x^{2}+7 x+1=0$ Solution: Given: $13 x^{2}+7 x+1=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=13, b=7$ and $c=1$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-7+\sqrt{49-4 \times 13 \times 1}}{2 \times 13}$ and $\beta=\frac{-7-\sqrt{49-4 \times 13 \times 1}}{2 \times 13}$ $\Rightarrow \alpha=\frac{-7+\sqrt{49-52}}{26} \quad$...
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Question: $8 x^{2}-9 x+3=0$ Solution: Given: $8 x^{2}-9 x+3=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=8, b=-9$ and $c=3$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{9+\sqrt{81-4 \times 8 \times 3}}{2 \times 8}$ and $\beta=\frac{9-\sqrt{81-4 \times 8 \times 3}}{2 \times 8}$ $\Rightarrow \alpha=\frac{9+\sqrt{81-96}}{16} \quad$ and $\qu...
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Question: $21 x^{2}-28 x+10=0$ Solution: Given: $21 x^{2}-28 x+10=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=21, b=-28$ and $c=10$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{28+\sqrt{784-4 \times 21 \times 10}}{2 \times 21}$ and $\beta=\frac{28-\sqrt{784-4 \times 21 \times 10}}{2 \times 21}$ $\Rightarrow \alpha=\frac{28+\sqrt{-56}}{4...
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Question: $17 x^{2}+28 x+12=0$ Solution: Given: $17 x^{2}+28 x+12=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=17, b=28$ and $c=12$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-28+\sqrt{784-4 \times 17 \times 12}}{34}$ and $\beta=\frac{-28-\sqrt{784-4 \times 17 \times 12}}{34}$ $\Rightarrow \alpha=\frac{-28+\sqrt{784-816}}{34} \quad$ an...
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Question: $27 x^{2}-10+1=0$ Solution: Given: $27 x^{2}-10 x+1=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=27, b=-10$ and $c=1$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{10+\sqrt{100-4 \times 27 \times 1}}{2 \times 27} \quad$ and $\quad \beta=\frac{10-\sqrt{100-4 \times 27 \times 1}}{2 \times 27}$ $\Rightarrow \alpha=\frac{10+\sqrt{10...
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Question: $17 x^{2}-8 x+1=0$ Solution: Given: $17 x^{2}-8 x+1=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=17, b=-8$ and $c=1$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{8+\sqrt{64-4 \times 17 \times 1}}{2 \times 17} \quad$ and $\quad \beta=\frac{8-\sqrt{64-4 \times 17 \times 1}}{2 \times 17}$ $\Rightarrow \alpha=\frac{8+\sqrt{64-68}}{...
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Question: $x^{2}+x+1=0$ Solution: We have: $x^{2}+x+1=0$ $\Rightarrow x^{2}+x+\frac{1}{4}+\frac{3}{4}=0$ $\Rightarrow\left(x+\frac{1}{2}\right)^{2}-\frac{3}{4} i^{2}=0$ $\Rightarrow\left(x+\frac{1}{2}\right)^{2}-\left(\frac{i \sqrt{3}}{2}\right)^{2}=0$ $\Rightarrow\left(x+\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left(x+\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$ $\Rightarrow\left(x+\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=0$ or $\left(x+\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$ $\Rightarrow x=-\fr...
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Question: $x^{2}-x+1=0$ Solution: We have: $x^{2}-x+1=0$ $\Rightarrow x^{2}-x+\frac{1}{2}+\frac{3}{4}=0$ $\Rightarrow x^{2}-2 \times x \times \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\frac{3}{4} i^{2}=0$ $\Rightarrow\left(x-\frac{1}{2}\right)^{2}-\left(\frac{i \sqrt{3}}{2}\right)^{2}=0$ $\Rightarrow\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$ $\Rightarrow\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=0 \quad$ or $\quad\left(x-\frac{1}{2}-\fra...
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[question] Question. If both $(x-2)$ and $\left(x-\frac{1}{2}\right)$ are factors of $p x^{2}+5 x+r$, prove that $p=r$. [/question] [solution] Solution: Let $f(x)=p x^{2}+5 x+r$ It is given that $(x-2)$ is a factor of $f(x)$. Using factor theorem, we have $f(2)=0$ $\Rightarrow p \times\left(\frac{1}{2}\right)^{2}+5 \times \frac{1}{2}+r=0$ $\Rightarrow \frac{p}{4}+r=-\frac{5}{2}$ $\Rightarrow p+4 r=-10 \quad \ldots \ldots(2)$ From (1) and (2), we have $4 p+r=p+4 r$ $\Rightarrow 4 p-p=4 r-r$ $\Rig...
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Question: $21 x^{2}+9 x+1=0$ Solution: Given: $21 x^{2}+9 x+1=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=21, b=9$ and $c=1$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-9+\sqrt{81-4 \times 21 \times 1}}{2 \times 21} \quad$ and $\quad \beta=\frac{-9-\sqrt{81-4 \times 21 \times 1}}{2 \times 21}$ $\Rightarrow \alpha=\frac{-9+\sqrt{3} i}{...
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Question: $5 x^{2}-6 x+2=0$ Solution: Given: $5 x^{2}-6 x+2=0$ Comparing the given equation with general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=5, b=-6$ and $c=2$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{6+\sqrt{36-4 \times 5 \times 2}}{2 \times 5}$ and $\beta=\frac{6-\sqrt{36-4 \times 2 \times 5}}{2 \times 5}$ $\Rightarrow \alpha=\frac{6+\sqrt{-4}}{10} \quad$ and $\quad \bet...
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Question: $x^{2}+2 x+5=0$ Solution: We have: $x^{2}+2 x+2=0$ $\Rightarrow x^{2}+2 x+1+1=0$ $\Rightarrow x^{2}+2 \times x \times 1+1^{2}-(i)^{2}=0$ $\Rightarrow(x+1)^{2}-(i)^{2}=0$ $\Rightarrow(x+1+i)(x+1-i)=0$ $\Rightarrow(x+1+i)=0$ or $(x+1-i)=0$ $\Rightarrow x=-1-i$ or $x=-1+i$ Hence, the roots of the equation are $-1+i$ and $-1-i$....
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Question: $x^{2}-4 x+7=0$ Solution: We have: $x^{2}-4 x+7=0$ $\Rightarrow x^{2}-4 x+4+3=0$ $\Rightarrow x^{2}-2 \times x \times 2+2^{2}-(\sqrt{3} i)^{2}=0$ $\Rightarrow(x-2)^{2}-(\sqrt{3} i)^{2}=0$ $\Rightarrow(x-2+\sqrt{3} i)(x-2-\sqrt{3} i)=0$ $\Rightarrow(x-2+\sqrt{3} i)=0$ or, $(x-2-\sqrt{3} i)=0$ $\Rightarrow x=2-\sqrt{3} i \quad$ or, $\quad x=2+\sqrt{3} i$ Hence, the roots of the equation are $2 \pm i \sqrt{3}$....
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Question: $4 x^{2}+1=0$ Solution: We have: $4 x^{2}+1=0$ $\Rightarrow(2 x)^{2}-i^{2}=0$ $\Rightarrow(2 x)^{2}-(i)^{2}=0$ $\Rightarrow(2 x+i)(2 x-i)=0$ $\Rightarrow(2 x+i)=0$ or $(2 x-i)=0$ $\Rightarrow 2 x=-i$ or $2 x=i$ $\Rightarrow x=-\frac{i}{2}$ or $x=\frac{i}{2}$ Hence, the roots of the equation are $\frac{1}{2} i$ and $-\frac{1}{2} i$....
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Question: $x^{2}+x+1=0$ Solution: We have: $x^{2}+x+1=0$ $\Rightarrow x^{2}+x+\frac{1}{4}+\frac{3}{4}=0$ $\Rightarrow x^{2}+\left(\frac{1}{2}\right)^{2}+2 \times x \times \frac{1}{2}-\left(\frac{\sqrt{3} i}{2}\right)^{2}=0$ $\Rightarrow\left(x+\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3} i}{2}\right)^{2}=0$ $\Rightarrow\left(x+\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\left(x+\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)=0$ $\Rightarrow\left(x+\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)=0$ or, $\left(x+\frac{...
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Question: $4 x^{2}-12 x+25=0$ Solution: We have: $4 x^{2}-12 x+25=0$ $\Rightarrow 4 x^{2}-12 x+9+16=0$ $\Rightarrow(2 x)^{2}+3^{2}-2 \times 2 x \times 3-(4 i)^{2}=0$ $\Rightarrow(2 x-3)^{2}-(4 i)^{2}=0$ $\Rightarrow(2 x-3+4 i)(2 x-3-4 i)=0 \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $\Rightarrow(2 x-3+4 i)=0$ or, $(2 x-3-4 i)=0$ $\Rightarrow 2 x=3-4 i$ or, $2 x=3+4 i$ $\Rightarrow x=\frac{3}{2}-2 i$ or, $x=\frac{3}{2}+2 i$ Hence, the roots of the equation are $\frac{3}{2}-2 i$ and $\frac{3}{2}+2 i...
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Question: $x^{2}+2 x+5=0$ Solution: Given: $x^{2}+2 x+5=0$ $x^{2}+2 x+5=0$ $\Rightarrow x^{2}+2 x+1+4=0$ $\Rightarrow(x+1)^{2}-(2 i)^{2}=0 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$ $\Rightarrow(x+1+2 i)(x+1-2 i)=0 \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $\Rightarrow(x+1+2 i)=0$ or, $(x+1-2 i)=0$ $\Rightarrow x=-(1+2 i)$ or, $x=-1+2 i$ Hence, the roots of the equation are $-1+2 i$ and $-1-2 i$....
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Question: $9 x^{2}+4=0$ Solution: Given: $9 x^{2}+4=0$ $9 x^{2}+4=0$ $\Rightarrow(3 x)^{2}+2^{2}=0$ $\Rightarrow(3 x)^{2}-(2 i)^{2}=0$ $\Rightarrow(3 x+2 i)(3 x-2 i)=0 \quad\left[\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right]$ $\Rightarrow(3 x+2 i)=0$ or, $(3 x-2 i)=0$ $\Rightarrow 3 x=-2 i$ or $3 x=2 i$ $\Rightarrow x=-\frac{2 i}{3}$ or $x=\frac{2 i}{3}$ Hence, the roots of the equation are $\frac{2 i}{3}$ and $-\frac{2 i}{3}$....
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Question: $x^{2}+1=0$ Solution: Given: $x^{2}+1=0$ $x^{2}+1=0$ $\Rightarrow x^{2}-1 i^{2}=0$ $\Rightarrow(x+i)(x-i)=0$ $\left[\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right]$ $\Rightarrow(x+i)=0$ or $(x-i)=0$ $\Rightarrow x=-i$ or $x=i$ Hence, the roots of the equation are $i$ and $-i$....
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Question: If $\alpha, \beta$ are roots of the equation $x^{2}-a(x+1)-c=0$, then write the value of $(1+\alpha)(1+\beta)$ Solution: Given: $x^{2}-a(x+1)-c=0$ or $x^{2}-a x-a-c=0$ Also, $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=-\left(\frac{-a}{1}\right)=a$ Product of the roots $=\alpha \beta=\frac{-(a+c)}{1}=-(a+c)$ $\therefore(1+\alpha)(1+\beta)=1+\beta+\alpha+\alpha \beta$ $=1+(\alpha+\beta)+(\alpha \beta)$ $=1+a-a-c$ $=1-c$...
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Question: If $\alpha, \beta$ are roots of the equation $x^{2}+l x+m=0$, write an equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$. Solution: Given equation: $x^{2}+l x+m=0$ Also, $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=\frac{-l}{1}=-l$ Product of the roots $=\alpha \beta=\frac{m}{1}=m$ Now, sum of the roots $=-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha \beta}=-\frac{-l}{m}=\frac{l}{m}$ Product of the roots $=\frac{1...
Read More →If α, β are roots of the equation x
Question: If $\alpha, \beta$ are roots of the equation $x^{2}+l x+m=0$, write an equation whose roots are $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$. Solution: Given equation: $x^{2}+l x+m=0$ Also, $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=\frac{-l}{1}=-l$ Product of the roots $=\alpha \beta=\frac{m}{1}=m$ Now, sum of the roots $=-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha \beta}=-\frac{-l}{m}=\frac{l}{m}$ Product of the roots $=\frac{1...
Read More →Write the number of quadratic equations, with real roots,
Question: Write the number of quadratic equations, with real roots, which do not change by squaring their roots. Solution: Let $\alpha$ and $\beta$ be the real roots of the quadratic equation $a x^{2}+b x+c=0$. On squaring these roots, we get: $\alpha=\alpha^{2} \quad$ and $\quad \beta=\beta^{2}$ $\Rightarrow \alpha(1-\alpha)=0$ and $\beta(1-\beta)=0$ $\Rightarrow \alpha=0, \alpha=1$ and $\beta=0,1$ Three cases arise: (i) $\alpha=0, \beta=0$ (ii) $\alpha=1, \beta=0$ (iii) $\alpha=1, \beta=1$ $(i...
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