Question

Without solving the following quadratic equations, find the value of $p$ for which the given equations have real and equal roots.
(i) $p{x}^2-4x+3=0$

(ii) $x^2+(p-3)x+p=0$

Answer 1

From equation (i),

${px}^2-4x+3=0$

$a=p,b=-4$ and $c=3$

If the roots of this equation are equal and real, then $D=0$.

Therefore,

$D=b^2-4ac=0$

${(-4)}^2-4\times p\times 3=0$

$16-12p=012p=16$

$p=\frac{4}{3}$

So, the value of p is $\frac{4}{3}.$

From equation (ii),

$x^2+(p-3)x+p=0$

$a=1,b=(p-3)$ and $c=p$

If the roots of this equation are equal and real, then $D=0$.

Therefore,

$D=b^2-4ac=0$

${(p-3)}^2-4\times 1\times p=0$

$p^2+9-6p-4p=0$

$p^2-10p+9=0$

$p^2-9p-p+9=0$

$p(p-9)-1(p-9)=0$

$(p-1)(p-9)$

$p=1,9$

So, the values of p is $1,9.$

Hence, this is correct answer.