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Question

(i) Find the value of k for which x = 1 is a root of the equation ${x}^{2}+kx+3=0$. Also, find the other root. (ii) Find the values of a and b for which $x=\frac{3}{4}\mathrm{and}x=-2$ are the roots of the equation $a{x}^{2}+bx-6=0.$

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Solution

(i) $\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\left(x=1\right)\text{is a root of (}{x}^{2}+kx+3=0\right).\phantom{\rule{0ex}{0ex}}\text{Therefore,}\left(x=1\right)\text{must satisfy the equation}\text{.}\phantom{\rule{0ex}{0ex}}⇒{\left(1\right)}^{2}+k×1+3=0\phantom{\rule{0ex}{0ex}}⇒\text{}k\text{+ 4 = 0}\phantom{\rule{0ex}{0ex}}⇒\text{}k=-4\phantom{\rule{0ex}{0ex}}\text{Hence, the required value of}k\text{is}-4\text{.}$ So, the equation becomes ${x}^{2}-4x+3=0$ On factorising we get; ${x}^{2}-x-3x+3=0\phantom{\rule{0ex}{0ex}}x\left(x-1\right)-3\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\left(x-1\right)\left(x-3\right)=0\phantom{\rule{0ex}{0ex}}⇒x-1=0\mathrm{or}x-3=0\phantom{\rule{0ex}{0ex}}⇒x=1\mathrm{or}x=3$ Hence, the other root is 3. (ii) $\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\frac{3}{4}\text{is a root of}a{x}^{2}+bx-6=0;\mathrm{therefore},\text{we have:}\phantom{\rule{0ex}{0ex}}a×{\left(\frac{3}{4}\right)}^{2}+b×\frac{3}{4}-6=0\phantom{\rule{0ex}{0ex}}⇒\frac{9a}{16}+\frac{3b}{4}=6\phantom{\rule{0ex}{0ex}}⇒\frac{9a+12b}{16}=6\phantom{\rule{0ex}{0ex}}⇒9a\text{+ 12}b-96=0\phantom{\rule{0ex}{0ex}}⇒\text{3}a\text{+ 4}b\text{= 32 ...}\left(\text{i}\right)\phantom{\rule{0ex}{0ex}}\text{Again, (}-2\right)\text{is a root of}a{x}^{2}+bx-6=0;\mathrm{therefore},\text{we have:}\phantom{\rule{0ex}{0ex}}a×{\left(-2\right)}^{2}+b×\left(-2\right)-6=0\phantom{\rule{0ex}{0ex}}⇒\text{4}a-2b=6\phantom{\rule{0ex}{0ex}}⇒\text{2}a-b=3\text{...}\left(\text{ii}\right)\phantom{\rule{0ex}{0ex}}\text{On multiplying}\left(\text{ii}\right)\text{by 4 and adding the result with}\left(\text{i}\right),\mathrm{w}\text{e get:}\phantom{\rule{0ex}{0ex}}⇒\text{3}a\text{+ 4}b\text{+ 8}a-\text{4}b\text{= 32 + 12}\phantom{\rule{0ex}{0ex}}⇒\text{11}a\text{= 44}\phantom{\rule{0ex}{0ex}}⇒a\text{= 4}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{in}\text{}\left(\text{ii}\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}2×4-b=3\phantom{\rule{0ex}{0ex}}⇒\text{8}-b\text{= 3}\phantom{\rule{0ex}{0ex}}⇒b\text{= 5}\phantom{\rule{0ex}{0ex}}\text{Hence, the required values of}a\text{and}b\text{are 4 and 5, respectively.}$

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