Question

The quadratic equation $x^2-7x+12=0$ can be simplified to , and it's corresponding roots will be .

• A. $0,1$
• B. $(x-\frac{1}{2}{)}^2=\frac{1}{4}$
• C. $(x-\frac{7}{2}{)}^2=\frac{1}{4}$
• D. $4,3$

Answer 1

Answer: (C), (D)

$4,3$

Given,
$x^2-7x+12=0$
Now, the equation can be written as:
$x^2-2\times \frac{7}{2}\times x+12=0\Rightarrow x^2-2\times \frac{7}{2}x+(\frac{7}{2}{)}^2-(\frac{7}{2}{)}^2+12=0\Rightarrow (x-\frac{7}{2}{)}^2-(\frac{7}{2}{)}^2+12=0\Rightarrow (x-\frac{7}{2}{)}^2=\frac{49}{4}-12=\frac{1}{4}\Rightarrow (x-\frac{7}{2}{)}^2=\frac{1}{4}$
Which is the simplified form of the given quadratic equation.
To find the roots, we now apply the square root on both sides.
$\Rightarrow x-\frac{7}{2}=\pm \frac{1}{2}\Rightarrow x-\frac{7}{2}=\frac{1}{2}\text{or}x-\frac{7}{2}=-\frac{1}{2}\Rightarrow x=4\text{or}x=3$