Question

find the zeroes of X2 +13x+1​

Answer 1

Factoring x2-13x-1

The first term is, x2 its coefficient is 1 .

The middle term is, -13x its coefficient is -13 .

The last term, "the constant", is -1

Step-1 : Multiply the coefficient of the first term by the constant 1 • -1 = -1

Step-2 : Find two factors of -1 whose sum equals the coefficient of the middle term, which is -13 .

-1 + 1 = 0

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Step-by-step explanation:

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=\frac{-13\pm\sqrt{165}}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}$

$\tt:\implies {x}^{2} + 13x + 1 = 0$

$\red{\underline \bold{To \: Find :}}$

$\tt:\implies Zeroes =?$

• Given Question

$\tt \circ \: a = 1 \: \: \: \: \: \: b = 13 \: \: \: \: \: \: c = 1$

$\bold{As \: we \: know \: that}$

$\tt: \implies D= {b}^{2} - 4ac$

$\tt: \implies D= {13}^{2} - 4 \times 1 \times 1$

$\tt: \implies D = 169 - 4$

$\green{\tt: \implies D = 165}$

$\bold{As \: we \: know \: that}$

$\tt: \implies x = \frac{ - b \pm \sqrt{D} }{2a}$

$\tt: \implies x = \frac{ - 13 \pm \sqrt{165} }{2 \times 1}$

$\green{\tt: \implies x = \frac{ - 13 \pm \sqrt{165} }{2} }$