Question

Let x be a real number, and x2-ax-21=(x+3)(x+b) with a, b constants . Find a and b.​

Answer 1

Ответ: a=4, b=-7.

Пошаговое объяснение:

x²-a*x-21=x²+x*(b+3)+3*b, -a*x-21=(b+3)*x+3*b ⇒

-a=b+3

3*b=-21 ⇒b=-7⇒a=4.

Answer 2

To find the values of constants 'a' and 'b' in the equation $x^2 - ax - 21 = (x + 3)(x + b)$, we can expand the right-hand side of the equation and then compare the coefficients of the corresponding powers of 'x'.

First, let's expand the right-hand side:

$(x + 3)(x + b) = x^2 + bx + 3x + 3b = x^2 + (b + 3)x + 3b$

Now, we can compare the coefficients of the terms on both sides of the equation:

1. Coefficient of $x^2$: On the left-hand side, the coefficient of $x^2$ is 1 (since the term is $x^2$). On the right-hand side, the coefficient of $x^2$ is also 1 (since the term is $x^2$). So, we have: $1 = 1$

2. Coefficient of $x$: On the left-hand side, the coefficient of $x$ is $-a$ (since the term is $-ax$). On the right-hand side, the coefficient of $x$ is $(b + 3)$. So, we have: $-a = b + 3$ ... (equation 1)

3. Constant term (coefficient of $x^0$): On the left-hand side, the constant term is $-21$. On the right-hand side, the constant term is $3b$. So, we have: $-21 = 3b$ ... (equation 2)

Now we have two equations (equation 1 and equation 2) with two unknowns (a and b). Let's solve them simultaneously:

From equation 2: $3b = -21$ Divide both sides by 3: $b = -21/3 = -7$

Now, substitute the value of 'b' into equation 1: $-a = b + 3$ $-a = -7 + 3$ $-a = -4$

Finally, multiply both sides by -1 to solve for 'a': $a = 4$

So, the values of 'a' and 'b' are: $a = 4$ and $b = -7$.