
How do you factor \[{x^4} - 61{x^2} + 900 = 0\] ?
Answer
553.8k+ views
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.
Complete step by step answer:
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,
Thus the factors are \[x = \pm 5\& x = \pm 6\].
This is our final answer.
Alternate method:
We also can find the factors by factoring the middle term such that the factors in addition give the middle term and the product gives the third term. The factors are -25 and -36 such that in addition they give -61 and on product it gives 900.
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.
Complete step by step answer:
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 \[x = \pm 5\& x = \pm 6\].
This is our final answer.
Alternate method:
We also can find the factors by factoring the middle term such that the factors in addition give the middle term and the product gives the third term. The factors are -25 and -36 such that in addition they give -61 and on product it gives 900.
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.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

