If the quadratic expression x2−(a−1)x+(a+14) were to be a perfect square then a=

Given expression x2−(a−1)x+(a+14)⇒ x2−2(a−12)x+(a−12)2−(a−12)2+(a+14)⇒ [x−(a−12)2]+(a+14)−(a−1)24To get perfect square Let (a+14)−(a−1)24=0⇒ a+14−(a2+1−2a)4=0⇒ a+14−a24−14+a2=0⇒ 3a2−a24=0⇒ a(32−a4)=0a=0, ⇒ 32−a4=0⇒ a4=32⇒ a= 32×4⇒ a=6∴ a=0, 6

If the quadratic expression x2−(a−1)x+(a+14) were to be a perfect square then a=

Given expression x2−(a−1)x+(a+14)⇒ x2−2(a−12)x+(a−12)2−(a−12)2+(a+14)⇒ [x−(a−12)2]+(a+14)−(a−1)24To get perfect square Let (a+14)−(a−1)24=0⇒ a+14−(a2+1−2a)4=0⇒ a+14−a24−14+a2=0⇒ 3a2−a24=0⇒ a(32−a4)=0a=0, ⇒ 32−a4=0⇒ a4=32⇒ a= 32×4⇒ a=6∴ a=0, 6

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$I f t h e q u a d r a t i c \exp r e s s i o n x^{2} - (a - 1) x + (a + \frac{1}{4}) w e r e t o b e a p e r f e c t s q u a r e t h e n a =$

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$0, 6$

2, 4

1, 8

4, 8

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n \exp r e s s i o n x^{2} - (a - 1) x + (a + \frac{1}{4}) \\ \Rightarrow x^{2} - 2 (\frac{a - 1}{2}) x + {(\frac{a - 1}{2})}^{2} - {(\frac{a - 1}{2})}^{2} + (a + \frac{1}{4}) \\ \Rightarrow [x - {(\frac{a - 1}{2})}^{2}] + (a + \frac{1}{4}) - \frac{{(a - 1)}^{2}}{4} \\ T o g e t p e r f e c t s q u a r e \\ L e t (a + \frac{1}{4}) - \frac{{(a - 1)}^{2}}{4} = 0 \\ \Rightarrow a + \frac{1}{4} - \frac{(a^{2} + 1 - 2 a)}{4} = 0 \\ \Rightarrow a + \frac{1}{4} - \frac{a^{2}}{4} - \frac{1}{4} + \frac{a}{2} = 0 \\ \Rightarrow \frac{3 a}{2} - \frac{a^{2}}{4} = 0 \\ \Rightarrow a (\frac{3}{2} - \frac{a}{4}) = 0 \\ a = 0, \Rightarrow \frac{3}{2} - \frac{a}{4} = 0 \\ \Rightarrow \frac{a}{4} = \frac{3}{2} \\ \Rightarrow a = \frac{3}{2} \times 4 \\ \Rightarrow a = 6 \\ ∴ a = 0, 6 \end{array}$