(i) Find the value of k for which x = 1 is a root of the equation
Question:

(i) Find the value of $k$ for which $x=1$ is a root of the equation $x^{2}+k x+3=0$. Also, find the other root.

(ii) Find the values of $a$ and $b$ for which $x=\frac{3}{4}$ and $x=-2$ are the roots of the equation $a x^{2}+b x-6=0$.

Solution:

(i)

It is given that $(x=1)$ is a root of $\left(x^{2}+k x+3=0\right)$.

Therefore, $(x=1)$ must satisfy the equation.

$\Rightarrow(1)^{2}+k \times 1+3=0$

$\Rightarrow k+4=0$

$\Rightarrow k=-4$

Hence, the required value of $k$ is $-4$.

So, the equation becomes $x^{2}-4 x+3=0$

On factorising we get;

$x^{2}-x-3 x+3=0$

$x(x-1)-3(x-1)=0$

$(x-1)(x-3)=0$

$\Rightarrow x-1=0$ or $x-3=0$

$\Rightarrow x=1$ or $x=3$

Hence, the other root is 3.

(ii)

It is given that $\frac{3}{4}$ is a root of $a x^{2}+b x-6=0$; therefore, we have:

$a \times\left(\frac{3}{4}\right)^{2}+b \times \frac{3}{4}-6=0$

$\Rightarrow \frac{9 a}{16}+\frac{3 b}{4}=6$

$\Rightarrow \frac{9 a+12 b}{16}=6$

$\Rightarrow 9 a+12 b-96=0$

$\Rightarrow 3 a+4 b=32 \quad \ldots$ (i)

Again, $(-2)$ is a root of $a x^{2}+b x-6=0 ;$ therefore, we have:

$a \times(-2)^{2}+b \times(-2)-6=0$

$\Rightarrow 4 a-2 b=6$

$\Rightarrow 2 a-b=3 \quad \ldots$ (ii)

On multiplying (ii) by 4 and adding the result with (i), we get:

$\Rightarrow 3 a+4 b+8 a-4 b=32+12$

$\Rightarrow 11 a=44$

$\Rightarrow a=4$

Putting the value of $a$ in (ii), we get:

$2 \times 4-b=3$

$\Rightarrow 8-b=3$

$\Rightarrow b=5$

Hence, the required values of $a$ and $b$ are 4 and 5 , respectively.