$D=(ab{)}^2+4a^2{b}^2=5a^2{b}^2$
$a^2{b}^2$has to be greater than or equal to 0 (i.e.,)$D>0.$So,roots are always real.
 

Let the roots be a,b.
$ab=k,a+b=8$
$a^2+b^2=30$
$(a+b{)}^2=64,a^2+b^2+2ab=64$
$30+2k=64,k=17$

Substitute x by (x-2)
The equation becomes:$(x-2{)}^2-3(x-2)+2=0,x^2-7x+12=0.$

Substitute x by$\left(\frac{x}{2}\right)$

The equation becomes:${\left(\frac{x}{2}\right)}^2-3\left(\frac{x}{2}\right)+2=0,x^2-6x+8=0.$ 

Soln:

Interchange ‘a’ and ‘c’ in the equation$a{x}^2+bx+c=0$

The equation becomes:$2x^2-3x+1=0.$ Hence option (a)

Sum of the roots:$p+q=-5$

Product of the roots:$pq=2$
$\frac{p}{q}+\frac{q}{p}=\frac{(p^2+q^2)}{pq}=\frac{\left[\right(p+q{)}^2-2pq]}{pq}=\frac{\left[\right(p+q{)}^2]}{pq}-2=\frac{25}{2}-2=\frac{21}{2}$

The equations$x^2-kx-21=0$and$x^2-3kx+35=0$have a common root.

So, equating the two equations:

$x^2-kx-21=x^2-3kx+35$

$2kx=56$

$k=\frac{28}{x}$

Putting in the equation:

$x^2-28-21=0$

$x^2=49$

$x=+7$

$ork=+4,As,k>0,k=4.$ 

For a quadratic equation$a{x}^2+bx+c=0$, if the roots are in ratio$m:n$then$mn{b}^2=(m+n{)}^{2}ac.$

In the equation given:

$4\times 3\times b^2=(4+3{)}^2(3\times 3)$

$12b^2=49\times 9$

$b=\frac{7\times 3}{\left(2\sqrt{3}\right)}=\frac{7\sqrt{3}}{2}$

For the equation given
$\text{Â}D=(k+5{)}^2+4(1\left)\right(k+5)=k^2+10k+25+4k+20=k^2+14k+45$

For the equation to have two distinct roots:
$\text{Â}D>0$

$k^2+14k+45>0\text{Â}(k+9)(k+5)>0$

Value of k will be$\text{Â}(−\mathrm{∞},−9)∪(−5,\mathrm{∞})$

 

Both the roots are greater than 1 and a$\text{Â}>\text{Â}$0. So, the graph of the equation will be:

 

 

$f\left(x\right)=a{x}^2+bx+c$

$f\left(1\right)>a+b+c>0$