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 $x=\frac{3}{4}$ and x=-2 are the roots of the equations $a{x}^2+bx-6=0$.

Answer 1

(i) $x^2+kx+3=0$
one root is given $x=1$
$(1{)}^2+k(1)+3=0$
$1+k+3=0$
$k=-4$

Now using this our quadratic equation would be

$x^2-4x+3=0$
$x^2-(3+1)x+3=0$
$x^2-3x-x+3=0$
$x(x-3)-1(x-3)=0$
$(x-3)(x-1)=0$
so our roots are
$x=3and1$

(ii) Given roots
$\alpha =\frac{3}{4}and\beta =-2$
$x^2-(\alpha +\beta )x+\alpha \beta$
$x^2-(\frac{3}{4}-2)x+\left(\frac{3}{4}\right)\times (-2)=0$
$x^2-\left(\frac{-5}{4}\right)x+\left(\frac{-3}{2}\right)=0$
$x^2+\frac{5}{4}x-\frac{3}{2}=0$
$4x^2+5x-6=0$
we have
$a{x}^2+bx-6=0$

so $a=4$ and $b=5$