In the given equation, [tex]x^2 + b x + c = 0[/tex], [tex]b[/tex] and [tex]c[/tex] are constants.

If [tex]-b + \sqrt{b^2 - 4c} = 18[/tex] and [tex]-b - \sqrt{b^2 - 4c} = 10[/tex], what is one possible value of [tex]x[/tex]?



Answer :

To determine one possible value of [tex]\( x \)[/tex] for the quadratic equation [tex]\( x^2 + bx + c = 0 \)[/tex], given the conditions [tex]\( -b + \sqrt{b^2 - 4c} = 18 \)[/tex] and [tex]\( -b - \sqrt{b^2 - 4c} = 10 \)[/tex], we can solve the problem step by step.

### Step 1: Set Up the Equations
We start with the given equations:
[tex]\[ -b + \sqrt{b^2 - 4c} = 18 \][/tex]
[tex]\[ -b - \sqrt{b^2 - 4c} = 10 \][/tex]

### Step 2: Add the Equations
Adding these two equations helps to eliminate the square root term:
[tex]\[ (-b + \sqrt{b^2 - 4c}) + (-b - \sqrt{b^2 - 4c}) = 18 + 10 \][/tex]
[tex]\[ -2b = 28 \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = -14 \][/tex]

### Step 3: Substitute [tex]\( b \)[/tex] Back into the Equations
We substitute [tex]\( b = -14 \)[/tex] back into one of the original equations to find [tex]\( \sqrt{b^2 - 4c} \)[/tex]:
[tex]\[ -(-14) + \sqrt{(-14)^2 - 4c} = 18 \][/tex]
[tex]\[ 14 + \sqrt{196 - 4c} = 18 \][/tex]
[tex]\[ \sqrt{196 - 4c} = 4 \][/tex]
Squaring both sides to remove the square root:
[tex]\[ 196 - 4c = 16 \][/tex]
Solving for [tex]\( c \)[/tex]:
[tex]\[ 4c = 180 \][/tex]
[tex]\[ c = 45 \][/tex]

### Step 4: Form the Quadratic Equation
Now that we have the values of [tex]\( b \)[/tex] and [tex]\( c \)[/tex]:
[tex]\[ b = -14 \][/tex]
[tex]\[ c = 45 \][/tex]

The quadratic equation becomes:
[tex]\[ x^2 - 14x + 45 = 0 \][/tex]

### Step 5: Solve the Quadratic Equation
To find [tex]\( x \)[/tex], we solve the quadratic equation:
[tex]\[ x^2 - 14x + 45 = 0 \][/tex]

We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, \; b = -14, \; c = 45 \][/tex]

[tex]\[ x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{14 \pm \sqrt{196 - 180}}{2} \][/tex]
[tex]\[ x = \frac{14 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{14 \pm 4}{2} \][/tex]

This gives us two solutions:
[tex]\[ x = \frac{14 + 4}{2} = 9 \][/tex]
[tex]\[ x = \frac{14 - 4}{2} = 5 \][/tex]

Therefore, one possible value of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{5} \][/tex]

And another possible value of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{9} \][/tex]