Question

x^2 - 29x + 120 = 0 where x is the denominator

Answer 1

≡QUESTION≡

Find the value of x in x² - 29x + 120 = 0, where x is the zero of the polynomial.

                                                           

║⊕ANSWER⊕║

Given equation,

$x^{2} -29x+120=0$

Solving, we get,

$x^{2} -29x+120=0\\x^{2} -24x-5x+120=0\\x(x-24)-5(x-24)=0\\(x-5)(x-24)=0$

Now,

$x-5=0\\=>x=5$

$x-24=0\\=> x=24$

∴ The values of x are 5 and 24

                                                         

Answer 2

Answer:

24,5

Step-by-step explanation:

${x}^{2} - 29x + 120 = 0 \\ \\ {x}^{2} - 5x - 24x + 120 = 0 \\ \\ x(x - 5) - 24(x - 5) = 0 \\ \\ (x - 24)(x - 5) = 0$

So, 24 and 5 are the zero of the given equation .