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# How do you factor ${x^4} - 61{x^2} + 900 = 0$ ?

Last updated date: 20th Jun 2024
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Hint:Given equation is a quadratic equation. But we will reconsider the terms so that they are in standard quadratic form. Then we will use a quadratic equation formula to find the roots or we can say to factorize the given expression. Here though the numbers are too large we can use this method simply to find the roots. But the equation is of the degree 4 so there will be 4 roots. They may be equal or unequal but there are 4 roots.

Given that,
${x^4} - 61{x^2} + 900 = 0$
Now we will write ${x^4}$ as ${\left( {{x^2}} \right)^2}$
So the equation becomes,
${\left( {{x^2}} \right)^2} - 61{x^2} + 900 = 0$
Now comparing with the general quadratic equation, $a = 1,b = - 61\& c = 900$
Putting these values in quadratic equation formula we get,
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - \left( { - 61} \right) \pm \sqrt {{{\left( { - 61} \right)}^2} - 4 \times 1 \times 900} }}{{2 \times 1}}$
On solving the brackets and root,
$= \dfrac{{61 \pm \sqrt {3721 - 3600} }}{2}$
Subtracting the numbers in root,
$= \dfrac{{61 \pm \sqrt {121} }}{2}$
Taking the square root,
$= \dfrac{{61 \pm 11}}{2}$
Now separating the roots we get,

 From quadratic formula $\dfrac{{61 + 11}}{2} =\dfrac{{72}}{2} = 36$ $\dfrac{{61 - 11}}{2} =\dfrac{{50}}{2} = 25$ Value of ${x^2}$ 36 $25$ Value of $x$ or roots of the equation $\pm 6$ $\pm 5$

Thus the factors are $x = \pm 5\& x = \pm 6$.
Note: Note that here we have written given equation ${x^4} - 61{x^2} + 900 = 0$ as ${\left( {{x^2}} \right)^2} - 61{x^2} + 900 = 0$ such that general quadratic equation is $a{x^2} + bx + c = 0$.thus in general the roots are equated to value of x. so here $x$ is nothing but ${x^2}$. And thus we have four roots of the given equation.