Question

If x is a positive integer such that the distance between points P(x, 2) and Q(3-6) is 10 units, then x =​

Answer 1

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Q1) If x is a positive integer such that the distance between points p(x,2) and Q(3,-6) is 10 units then x=

$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \: \large \orange{ \bold{ \underline{ \underline{step \: by \: step \: explaination}}}}$

$\: \quad\implies \red { \underline{ \displaystyle \bf{ by \: distance \: formula}}}$

$\: \\ \quad \implies \displaystyle \sf{ pq = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }$

$\: \: \\ \quad \implies \displaystyle \sf{10units = \sqrt{ {(3 - x)}^{2} + {(6 - 2)}^{2} }}$

$\: \quad \implies \displaystyle \sf{10 = \sqrt{ {(3 - x)}^{2} + 16}}$

$\: \quad \implies \displaystyle \sf{100 = {(x - 3)}^{2} + 16}$

$\: \: \quad \implies \displaystyle \sf{100 - 16 = {(x - 3)}^{2} }$

$\: \: \quad \implies \displaystyle \sf{84 = {x}^{2} + 9 - 6x}$

$\: \: \quad \implies \displaystyle \sf{ {x}^{2} - 6x + 9 = 84}$

$\: \: \quad \implies \displaystyle \sf{ {x}^{2} - 6x = 84 - 9}$

$\: \: \quad \implies \displaystyle \sf{ {x}^{2} - 6x = 75}$

$\: \\ \quad \implies \displaystyle \bf{quadratic \ \: equation : {x}^{2} - 6x - 75 = 0}$

$\: \: \\ \quad \implies \displaystyle \red{\bf{ \underline{ \: by \: determinant \: form}}}$

$\: \: \quad \implies \displaystyle \sf{ {b}^{2} - 4ac}$

$\: \: \quad\implies \displaystyle \sf{ {( - 6)}^{2} - 4(1)( - 75)}$

$\: \quad \implies \displaystyle \sf{36 + 300}$

$\: \quad\implies \displaystyle \sf{336}$

$\: \: \quad \implies \red{ \bf{ \underline{by \: formula \: method}}}$

$\: \quad \implies \displaystyle \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\: \quad\implies \displaystyle \sf{x = \frac{6 \pm \sqrt{336} }{2} }$

$\: \quad \implies \displaystyle \sf{x = \frac{6 \pm \sqrt{36 + 300} }{2} }$

$\: \: \quad \implies \displaystyle \sf{x = \frac{6 \pm6 \sqrt{300} }{2} }$

$\: \quad\implies \displaystyle \sf{x = \frac{2(3 \pm3 \sqrt{300}) }{2} }$

$\: \: \quad \implies \displaystyle \sf{x = 3 \pm3 \sqrt{300} }$

$\: \red \bigstar \boxed{ \boxed{ \underline{ \underline{ \tt{ \: the \: roots \: are \: 3 + 3 \sqrt{300} \: and \: 3 - 3 \sqrt{300} }}}}}$