Question

$-x^2 + bx - 169 = 0$
in the given equation, $b$ is an integer. the equation has no real solutions. what is the least possible value of $b$?

Explanation:

Step1: Rewrite equation in standard form

Multiply by -1: $x^2 - bx + 169 = 0$

Step2: Use discriminant condition for no real roots

Discriminant $\Delta = b^2 - 4ac < 0$. Here $a=1, c=169$, so $\Delta = b^2 - 4(1)(169) < 0$

Step3: Solve inequality for b

$b^2 < 676 \Rightarrow |b| < 26 \Rightarrow -26 < b < 26$

Step4: Find least integer b

Since b is integer, least possible value is -25 (wait correction: wait discriminant for original equation $-x^2 + bx -169=0$: $a=-1, b=b, c=-169$, $\Delta = b^2 - 4(-1)(-169)=b^2 - 676 <0$ same result. So b must satisfy $b^2 <676$. The integers b are from -25 to 25. Least possible integer is -25.

Wait correction: Wait $b^2 <676$ means $-26

Final answer: -25? Wait wait no wait: wait the question says "least possible value of b". Integers less than 26 and greater than -26: the smallest integer is -25. Yes.

Wait wait initial mistake in step3 sign: $b^2 <676$ so $-26

Step1: Identify quadratic form

Original equation: $-x^2 + bx -169=0$

Step2: Calculate discriminant

Discriminant $\Delta = b^2 - 4ac$ where $a=-1, c=-169$: $\Delta = b^2 - 4(-1)(-169)=b^2 - 676$

Step3: Set discriminant <0 for no real roots

$b^2 -676 <0 \Rightarrow b^2 <676$

Step4: Determine integer b range

$-26

Step5: Find least integer b

Least integer in range is -25.

Answer:

27