Question

the zeros of the quardratic polynomial x^2 +24x+119​

Answer 1

Answer:

${(x)}^{2} + 24x \: + 119 \\ \\ {(x)}^{2} + 17x \: + 7x \: + 119 \\ \\ x {}(x \: + 17) \: 7(x + 17) \\ \\ (x + 7) \: (x + 17)$

( X + 7 ). ( X + 17 )

X = -7

X = -17

Answer 2

Answer:

$\text{The Correct Answer is x=-7 \ and \ x=-17 }$

Step-by-step explanation:

We have to calculate zeros of the quadratic polynomial

$x^2+24x+119=0$

$\text{Solution : }\\\text{Using the quadratic formula where a=1,b=24,c=119 }\\then\ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-24\pm\sqrt{24^2-4(1)(119)}}{2(1)} \\ x=\frac{-24\pm\sqrt{576-476}}{2} \\ x=\frac{-24\pm\sqrt{100}}{2} \\\\\text{The Discriminant b^2-4ac > 0 so, there are two real roots.}$

Simplify the radical:

$x=\frac{-24\pm 10}{2} \\ x=-\frac{14}{2} \ \ \ x=-\frac{34}{2} \\\text{which becomes}\\\\ x=-7 and x=-17$

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