Question

Find the square root of $x^4-6x^3+19x^2-30x+25$.

• A. $|x^2-3x+5|$.
• B. $|x^2+3x-5|$.
• C. $|x^2+3x+5|$.
• D. None of these

Answer 1

The correct option is A $|x^2-3x+5|$.

To find the square root of the given polynomial $x^4-6x^3+19x^2-30x+25$, we must equate it to the general form of equation that is $(a{x}^2+bx+c)$ as shown below:

$\sqrt{x^4-6x^3+19x^2-30x+25}=a{x}^2+bx+c\Rightarrow {\left(\sqrt{x^4-6x^3+19x^2-30x+25}\right)}^2={(a{x}^2+bx+c)}^2\Rightarrow x^4-6x^3+19x^2-30x+25={(a{x}^2+bx+c)}^2\Rightarrow x^4-6x^3+19x^2-30x+25={\left(a{x}^2\right)}^2+{\left(bx\right)}^2+{\left(c\right)}^2+(2\times a{x}^2\times bx)+(2\times bx\times c)+(2\times c\times a{x}^2)(\because (a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca)\Rightarrow x^4-6x^3+19x^2-30x+25=a^2{x}^4+b^2{x}^2+c^2+2ab{x}^3+2bcx+2ac{x}^2$

Now, comparing the coefficients, we get:

$a^2=1,b^2+2ac=19,c^2=25,2ab=-6,2bc=-30$

$a^2=1\Rightarrow a=1$

$c^2=25\Rightarrow c=5$

$2ab=-6\Rightarrow 2\times 1\times b=-6\Rightarrow 2b=-6\Rightarrow b=-3$

Therefore, $a=1,b=-3,c=5$ and substituting the values in the equation $(a{x}^2+bx+c)$, we get $x^2-3x+5$.

Hence, the square root of $x^4-6x^3+19x^2-30x+25$ is $\left|\right(x^2-3x+5\left)\right|$.