Question
Determine whether the given quadratic equations have equal roots and if so, find the roots:
$\frac{4}{3} x^2-2 x+\frac{3}{4}=0$
$\frac{4}{3} x^2-2 x+\frac{3}{4}=0$
✓
Answer
We have
$\frac{4}{3} x^2-2 x+\frac{3}{4}=0$
Here, $a=\frac{4}{3}, b=-2$ and $c=\frac{3}{4}$
Discriminant
$=b^2-4 a c$
$ =(-2)^2-4 \times \frac{4}{3} \times \frac{3}{4} $
$ =4-4 $
$ =0$
So, the given equation has two real and equal roots given by
$a=\frac{-b+\sqrt{b^2-4 a c}}{2 a}$
$=\frac{+2+0}{4}$
$=\frac{3}{4}$
and $\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$
$=\frac{+2-0}{2 \times \frac{4}{3}} $$=\frac{3}{4} .$
$\frac{4}{3} x^2-2 x+\frac{3}{4}=0$
Here, $a=\frac{4}{3}, b=-2$ and $c=\frac{3}{4}$
Discriminant
$=b^2-4 a c$
$ =(-2)^2-4 \times \frac{4}{3} \times \frac{3}{4} $
$ =4-4 $
$ =0$
So, the given equation has two real and equal roots given by
$a=\frac{-b+\sqrt{b^2-4 a c}}{2 a}$
$=\frac{+2+0}{4}$
$=\frac{3}{4}$
and $\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$
$=\frac{+2-0}{2 \times \frac{4}{3}} $$=\frac{3}{4} .$
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