Question

find the zeroes of X2 +13x+1​

Answer 1

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

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

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

$\green{\underline \bold{Given :}} \\ \tt:\implies {x}^{2} + 13x + 1 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt:\implies Zeroes =?$

• According to given question :

$\tt \circ \: a = 1 \: \: \: \: \: \: b = 13 \: \: \: \: \: \: c = 1 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies D= {b}^{2} - 4ac \\ \\ \tt: \implies D= {13}^{2} - 4 \times 1 \times 1 \\ \\ \tt: \implies D = 169 - 4 \\ \\ \green{\tt: \implies D = 165} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \tt: \implies x = \frac{ - 13 \pm \sqrt{165} }{2 \times 1} \\ \\ \green{\tt: \implies x = \frac{ - 13 \pm \sqrt{165} }{2} }$

Answer 2

$\mathbb{\underline{\underline{Question}}}$

$\tt\red{find\: the\:zeroes\: of\: x^2\:+\:13x\:+\:1}$

$\mathbb{\underline{\underline{To\:find}}}$

$\tt\pink{Value\:of\:0}$

$\mathbb{\underline{\underline{Solution}}}$

$\tt{As\:to\:question,}$

◦a = 1

◦b = 13

◦c = 1

$\tt{As\:we\:know,}$

$\mathtt{D=b^2-4ac}$

⟹ $\mathtt{D=13^2-4×1×1}$

⟹$\tt{D=169-4}$

⟹$\tt{D=165}$

$\tt{we\:know\:that,}$

⟹$\tt{x}$ = $\frac{-b+-rootD}{2a}$

⟹$\tt{x}$ = $\frac{-13+-root165}{2}$

⟹$\tt{x}$ = $\frac{-13+-root165}{2}$