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# Show that the equation ${x^4} - 5{x^3} + 3{x^2} + 35x - 70 = 0$ has a root between 2 and 3 and one between -2 and -3.

Last updated date: 13th Jun 2024
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Hint: To approach this solution to substitute the given values in the given equation and show that at the given equation have roots between the given interval, using this information will help you to approach the solution.

Complete step-by-step solution:
With the help of a sign of function value, we will get the information about unsolved roots.
Given equation ${x^4} - 5{x^3} + 3{x^2} + 35x - 70 = 0$
Consider $f\left( x \right) = {x^4} - 5{x^3} + 3{x^2} + 35x - 70$
Then
$f\left( 2 \right) = {2^4} - 5 \times {2^3} + 3 \times {2^2} + 35 \times 2 - 70$
$f\left( 2 \right) = - 12$
And
$f\left( 3 \right) = {3^4} - 5 \times {3^3} + 3 \times {3^2} + 35 \times 3 - 70$
$f\left( 3 \right) = 8$
Since the signs of $f\left( 2 \right)$ and $f\left( 3 \right)$ are opposite,
$f\left( x \right)$ must cross x-axis at least once in the interval $\left( {2,3} \right)$
Therefore, $f\left( x \right) = 0$ must have one root between 2 and 3.
Similarly,
$f\left( { - 2} \right) = {\left( { - 2} \right)^4} - 5{\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} + 35\left( { - 2} \right) - 70$
$f\left( { - 2} \right) = - 72$
And
$f\left( { - 3} \right) = {\left( { - 3} \right)^4} - 5{\left( { - 3} \right)^3} + 3{\left( { - 3} \right)^2} + 35\left( { - 3} \right) - 70$
$f\left( { - 3} \right) = 68$
Since the signs of $f\left( { - 2} \right)$and $f\left( { - 3} \right)$ are opposite.
Therefore, $f\left( x \right) = 0$ must have one root between -2 and -3.

Note: This question could have been solved by finding out all the four roots of the equation and then checking for the desired result. But the process done above is the easiest to solve as we don’t need to find all the roots, we just need to find the location of the roots.