Question

The integer value of x for which x² + 19x + 92 is square of an integer are

• A. – 8 & – 11
• B. – 8 & 11
• C. 8 & 11
• D. ± 8, ±

Answer 1

Answer: (A)

– 8 & – 11

Answer 2

Let x² + 19x + 92 = m² , m∈I

x² + 19x + 92 – m² = 0

x = $\frac{–19 \pm \sqrt{4m^{2} - 7}}{2}$

∴ x∈I

So, –19 ± $\sqrt{4m^{2} - 7}$ must be even integer

∴ $\sqrt{4m^{2} - 7}$ must be odd integer

Let$\sqrt{4m^{2} - 7}$ = 2 k + 1 ; k∈I

∴ 4m² – (2k + 1)² = 7

(2m – 2k – 1) (2m + 2k + 1) = 7

∴ 7 is prime

so possibilities are

case

(i) 2m–2k–1=1& 2m + 2k + 1 = 7 then m = 2

(ii) 2m–2k–1 = 7 & 2m + 2k + 1 = 1 then m = 2

at m = 2 ⇒ x = –8, –11

(iii) 2m – 2k –1 = – 1 & 2m + 2k + 1 = –7

then m = – 2

(iv) 2m–2k–1= –7 & 2m + 2k + 1= –1 then m = –2

at m = – 2 x = – 8, – 11