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 $a{x}^2+bx-6=0.$

Answer 1

(i)
$$\begin{gathered} \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}(x=1)\text{is a root of (}{x}^2+kx+3=0). \\ \text{Therefore,}(x=1)\text{must satisfy the equation}\text{.} \\ \Rightarrow {\left(1\right)}^2+k\times 1+3=0 \\ \Rightarrow k\text{+ 4 = 0} \\ \Rightarrow k=-4 \\ \text{Hence, the required value of}k\text{is}-4\text{.} \end{gathered}$$
So, the equation becomes $x^2-4x+3=0$
On factorising we get;
$$\begin{gathered} x^2-x-3x+3=0 \\ x(x-1)-3(x-1)=0 \\ (x-1)(x-3)=0 \\ \Rightarrow x-1=0\mathrm{or}x-3=0 \\ \Rightarrow x=1\mathrm{or}x=3 \end{gathered}$$
Hence, the other root is 3.

(ii)
$$\begin{gathered} \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\frac{3}{4}\text{is a root of}a{x}^2+bx-6=0;\mathrm{therefore},\text{we have:} \\ a\times {\left(\frac{3}{4}\right)}^2+b\times \frac{3}{4}-6=0 \\ \Rightarrow \frac{9a}{16}+\frac{3b}{4}=6 \\ \Rightarrow \frac{9a+12b}{16}=6 \\ \Rightarrow 9a\text{+ 12}b-96=0 \\ \Rightarrow \text{3}a\text{+ 4}b\text{= 32 ...}\left(\text{i}\right) \\ \text{Again, (}-2)\text{is a root of}a{x}^2+bx-6=0;\mathrm{therefore},\text{we have:} \\ a\times {(-2)}^2+b\times (-2)-6=0 \\ \Rightarrow \text{4}a-2b=6 \\ \Rightarrow \text{2}a-b=3\text{...}\left(\text{ii}\right) \\ \text{On multiplying}\left(\text{ii}\right)\text{by 4 and adding the result with}\left(\text{i}\right),\mathrm{w}\text{e get:} \\ \Rightarrow \text{3}a\text{+ 4}b\text{+ 8}a-\text{4}b\text{= 32 + 12} \\ \Rightarrow \text{11}a\text{= 44} \\ \Rightarrow a\text{= 4} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{in}\left(\text{ii}\right),\mathrm{we}\mathrm{get}: \\ 2\times 4-b=3 \\ \Rightarrow \text{8}-b\text{= 3} \\ \Rightarrow b\text{= 5} \\ \text{Hence, the required values of}a\text{and}b\text{are 4 and 5, respectively.} \end{gathered}$$