Consider the equation x2-x-a-0- a- N- If equation has integral roots then prove that a can be 2- 6- 12 and 20-

x^2+x a=0 x= 1±√(1+4a)2 #160; 1+4a must be #160; a perfect square integerSo a can be 2,6,12,20 ...

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Standard X

Mathematics

Completing the Square

Consider the ...

Question

Consider the equation $x^{2} + x - a = 0, a \in N$ . If equation has integral roots then prove that a can be $2, 6, 12$ and $20$ .

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Solution

$x^{2} + x - a = 0 ⟹ x = \frac{- 1 \pm \sqrt{1 + 4 a}}{2}$
$⟹ 1 + 4 a$ must be a perfect square integer
So $a$ can be $2, 6, 12, 20$

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Consider the equation x2+x−a=0,a∈N. If equation has integral roots then prove that a can be 2,6,12 and 20.

x2+x−a=0⟹x=−1±√1+4a2 ⟹1+4a must be a perfect square integerSo a can be 2,6,12,20

Consider the equation $x^{2} + x - a = 0, a \in N$ . If equation has integral roots then prove that a can be $2, 6, 12$ and $20$ .

$x^{2} + x - a = 0 ⟹ x = \frac{- 1 \pm \sqrt{1 + 4 a}}{2}$
$⟹ 1 + 4 a$ must be a perfect square integer
So $a$ can be $2, 6, 12, 20$