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- From the equation whose roots are
- 3,-2
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Solution :1a
- -5,4
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Solution :1b
![Rendered by QuickLaTeX.com \sqrt{3},-\sqrt{3}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-5216f4fb22879848aeb3a723f53329c9_l3.png)
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Solution :1c
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Solution :1d
- -3+5i,-i-5i
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Solution :1e
- a+ib,a-ib
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Solution :1f
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- Find a quadratic equation whose roots are twice the roots of
![Rendered by QuickLaTeX.com 4x^2+8x-5=0](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-76f6e9649934cefb355a39c23c592cd9_l3.png)
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Solution :2a
- Find a quadratic equation whose roots are reciprocals of the roots of
![Rendered by QuickLaTeX.com 3x^2-5x-2=0](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-2d8cb9b6ee0c48afd094bcae82b735af_l3.png)
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Solution :2(b)
- Find a quadratic equation whose roots are greater by h than the roots of
![Rendered by QuickLaTeX.com x^2-px+q=0](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-6cda720db6b0b0693b6f6c86dc01d701_l3.png)
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Solution :2(c)
- Find a quadratic equation whose roots are the squares of the roots of
![Rendered by QuickLaTeX.com 3x^2-5x-2=0](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-2d8cb9b6ee0c48afd094bcae82b735af_l3.png)
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Solution :2(d)
- Find a quadratic equation with rational coefficients one of whose roots is
- 4+3i
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Solution :3a
-
![Rendered by QuickLaTeX.com \frac{1}{5+3i}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-6ff8224bddfbfb99b7ad102ff6a08509_l3.png)
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Solution :3b
![Rendered by QuickLaTeX.com 2+\sqrt{3}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-2c8f2a150c4ac53b66c42a8581afcd99_l3.png)
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Solution :3c
- Find the value of k so that the equation
has one root 3
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Solution :4a
has roots whose sum is equal to 6
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Solution :4b
has roots equal but opposite in sign
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Solution :4c
has roots numerically equal but opposite in sign
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Solution :4d
has one root equal to zero
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Solution :4e
has the reciprocal roots
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Solution :4f
has roots whose difference is ![Rendered by QuickLaTeX.com \frac{1}{4}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-83dc4a903812856a369855508d0b2637_l3.png)
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Solution :4g
- Show that -1 is a root of the equation
x^2+(2a-b-c)x+(c+a-2b)=0\). Find the other root.
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Solution :5
- Find the value of m for which the equation
will have (a) reciprocal roots (b) one root zero.
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Solution :6
- If the roots of the equation
differ by 1, prove that ![Rendered by QuickLaTeX.com a^2=4c+1](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-cb63f82dfb218d7f13d6fd5b6bbe5f3b_l3.png)
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Solution :7
- If
are the roots of the equation
, find the equation whose roots are
and ![Rendered by QuickLaTeX.com \beta ^2 \alpha ^{-1}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-67e5b899a3cc8a4b14ccac730d1b4375_l3.png)
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Solution :8a
and ![Rendered by QuickLaTeX.com \beta + \frac{1}{\alpha}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-bfb93cb41b4f721ed133aa0a6cf1d3fa_l3.png)
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Solution :8(b)
- If
are the roots of the equation
, find the equation whose roots are
and ![Rendered by QuickLaTeX.com \beta ^2 \alpha ^{-1}](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-67e5b899a3cc8a4b14ccac730d1b4375_l3.png)
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Solution :9a
and ![Rendered by QuickLaTeX.com \beta ^3](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-6b8243fe58560ba2a7576e6f3fa0922b_l3.png)
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Solution :9b
and ![Rendered by QuickLaTeX.com (\alpha+\beta)^2](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-b756fea4b3683986ee373c6345f4fa53_l3.png)
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Solution :9c
- the reciprocal of the roots of given equation
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Solution :9d
-
- If the roots of the equation
be in the ratio of 3:4, prove that ![Rendered by QuickLaTeX.com 12b^2=49ac](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-1bc7b02fd4d7d70cf69d344ddfa2f478_l3.png)
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Solution :10a
- If one root of the equation
be four times the other root, show that ![Rendered by QuickLaTeX.com 4b^2=25ac](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-107b00d7521687e3b327dc55c667d50d_l3.png)
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Solution :10b
- For what values of m, the equation
may have its root in the ratio 2:3
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Solution :10c
-
- If
are the roots of the equation
, prove that ![Rendered by QuickLaTeX.com \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-7b238a006900655c82e127e710bbf077_l3.png)
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Solution :11a
- If roots of the equation
be in the ratio of p:q, prove that ![Rendered by QuickLaTeX.com \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-7dd9e017419ca13d6dcfd073e1debd9c_l3.png)
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Solution :11b
- If one root of the equation
be square of the other root, prove that ![Rendered by QuickLaTeX.com b^3+a^2c+ac^2=3abc](https://mean.edu.np/wp-content/ql-cache/quicklatex.com-2b363d7154bd3c962a6b6bbdabdb0c9c_l3.png)
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Solution :12