Question

Arrange the following equations based on the descending order of their root values.(Use quadratic formula )

• A. $x^2-6x+9=0$
• B. $x^2-2x+1=0$
• C. $x^2-4x+4=0$
• D. $x^2+2x+1=0$

Answer 1

Let's solve the given equations using quadratic formula .

Quadratic formula :

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

1) $x^2-6x+9=0$

Using formula, we have

$x=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(1\left)\right(9)}}{2\left(1\right)}$

$\Rightarrow$$x=\frac{6\pm \sqrt{36-36}}{2}$
$\Rightarrow$$x=\frac{6\pm \sqrt{0}}{2}$
$\Rightarrow$$x=3$

2) $x^2-2x+1=0$

Using formula, we have

$x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(1\left)\right(1)}}{2\left(1\right)}$

$\Rightarrow$$x=\frac{2\pm \sqrt{4-4}}{2}$
$\Rightarrow$$x=\frac{2\pm \sqrt{0}}{2}$
$\Rightarrow$$x=1$

3) $x^2-4x+4=0$

Using formula, we have

$x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1\left)\right(4)}}{2\left(1\right)}$

$\Rightarrow$$x=\frac{4\pm \sqrt{16-16}}{2}$
$\Rightarrow$$x=\frac{4\pm \sqrt{0}}{2}$
$\Rightarrow$$x=2$

4) $x^2+2x+1=0$

Using formula, we have

$x=\frac{-\left(2\right)\pm \sqrt{(2{)}^2-4(1\left)\right(1)}}{2\left(1\right)}$

$\Rightarrow$$x=\frac{-2\pm \sqrt{4-4}}{2}$
$\Rightarrow$$x=\frac{-2\pm \sqrt{0}}{2}$
$\Rightarrow$$x=-1$