Question

Let P(x) = x2+ ax + 1. If P(x) is a negative integer for only one real x, then product of all values of 'a' is

Answer 1

As per the data given in the equation.

We have to find the product of all values of 'a'

Given

$P(x) = x²+ ax +1$

Step-by-step explanation:

Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax² + bx + c = 0 where a, b, c, ∈ R and a ≠ 0.

Now,

$P(x) = x²+ ax +1 \: \: \: \: \: ........(1)$

Only one root is there .

The roots of the quadratic equation:

$D = b²– 4ac$

$D=0$

$b²-4ac=0 \: \: \: \: \: \: \: \: \: \: ....(2)$

Compare the equation (1) by ax²+bx+c=0

The values are

$a=1 ,b= a \: and \: c=1$

Put the values in equation (2)

$a²-4×1×1=0$

$a²-4=0$

Shift the value 4 to the right side ,

$a²=4$

$a=√4$

$a=±2$

Hence,

The values are +2 and -2 .

And

The product of the values is

$+2×-2= -4$

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Answer 2

Answer:

-4 is the value of a.

Step-by-step explanation:

Explanation:

Given, P(x) = $x ^2 + ax + 1$.

Discriminant formula -The section of the quadratic formula following the square root symbol, $b^2 - 4ac$, is the discriminant. If there are two solutions, one solution, or none at all, the discriminant informs us.

Step 1:

Discriminant = $b^2 - 4ac$

⇒$b^2 - 4ac = 0$

where from the question we have, a = 1 , b = a  and c = 1.

On putting all these value in the discriminant formula we get,

$a^2 - 4(1)(1)$ = 0

⇒$a^2 = 4$ ⇒ a = $\sqrt{4}$ = $\frac{+}{}(2)$

Therefore, a = +2  and a = 2.

Now, the product of values of 'a' = -2 × 2 = -4

Final answer:

Hence, the value of a is -4 .

#SPJ3