If (b – 6) is one root of the quadratic equation x 2– 6x + b = 0, where b is an integer, then what is the maximum value of b 2?
81
The problem provides a quadratic equation, \(x^2 – 6x + b = 0\), and states that one of its roots is \((b – 6)\). We are also told that \(b\) is an integer. Our goal is to find the maximum possible value of \(b^2\).
In a quadratic equation \(Ax^2 + Bx + C = 0\), if \(x_0\) is a root, then substituting \(x_0\) into the equation will satisfy it, meaning \(Ax_0^2 + Bx_0 + C = 0\). We can use this property to find the value(s) of \(b\).
The given equation is \(x^2 - 6x + b = 0\). One root is given as \(x = b - 6\). Let's substitute this root into the equation:
\((b - 6)^2 - 6(b - 6) + b = 0\)
Now, we need to expand and simplify this equation to solve for \(b\).
Expand the equation:
\((b^2 - 12b + 36) - (6b - 36) + b = 0\)
Remove the parentheses and combine like terms:
\(b^2 - 12b + 36 - 6b + 36 + b = 0\)
\(b^2 + (-12b - 6b + b) + (36 + 36) = 0\)
\(b^2 - 17b + 72 = 0\)
This is a quadratic equation in terms of \(b\). We can solve for \(b\) by factoring, using the quadratic formula, or other methods. Factoring is suitable here as we need two numbers that multiply to 72 and add up to -17. These numbers are -8 and -9.
So, we can factor the equation as:
\((b - 8)(b - 9) = 0\)
This equation holds true if either factor is equal to zero.
Since the problem states that \(b\) is an integer, both \(b=8\) and \(b=9\) are valid possible values for \(b\).
We need to find the maximum value of \(b^2\). Let's calculate \(b^2\) for each valid value of \(b\):
Comparing the two possible values for \(b^2\), which are 64 and 81, the maximum value is 81.
Based on our analysis of the quadratic equation and its given root, the possible integer values for \(b\) are 8 and 9. Calculating \(b^2\) for these values gives us 64 and 81, respectively. The maximum among these is 81.
| Concept | Description | Application in Problem |
|---|---|---|
| Root of an Equation | A value of the variable that satisfies the equation when substituted. | Used to substitute \((b-6)\) for \(x\) in the quadratic equation. |
| Quadratic Equation | An equation of the form \(Ax^2 + Bx + C = 0\), where \(A \ne 0\). | The given equation \(x^2 - 6x + b = 0\) is a quadratic equation. The equation derived for \(b\), \(b^2 - 17b + 72 = 0\), is also a quadratic equation. |
| Factoring Quadratics | Expressing a quadratic polynomial as a product of linear factors. | Used to solve \(b^2 - 17b + 72 = 0\) for \(b\). |
| Integer | A whole number (positive, negative, or zero). | The problem specifies that \(b\) must be an integer, validating our solutions \(b=8\) and \(b=9\). |
For a general quadratic equation \(Ax^2 + Bx + C = 0\) with roots \(\alpha\) and \(\beta\), there are some useful properties (Vieta's formulas):
In our given equation \(x^2 - 6x + b = 0\), \(A=1\), \(B=-6\), and \(C=b\). Let the other root be \(\alpha\). The roots are \((b-6)\) and \(\alpha\).
Using the sum of roots property:
\((b-6) + \alpha = -\frac{-6}{1} = 6\)
\(\alpha = 6 - (b-6) = 6 - b + 6 = 12 - b\)
So, the two roots are \((b-6)\) and \((12-b)\).
Using the product of roots property:
\((b-6) \times (12-b) = \frac{b}{1} = b\)
Expand the left side:
\(12b - b^2 - 72 + 6b = b\)
\(-b^2 + 18b - 72 = b\)
Move all terms to one side to form a quadratic equation in \(b\):
\(0 = b^2 - 18b + 72 + b\)
\(b^2 - 17b + 72 = 0\)
This is the exact same quadratic equation for \(b\) that we obtained by substituting the root directly. Solving this equation again gives \(b=8\) or \(b=9\), leading to the same maximum value of \(b^2\) which is 81. This demonstrates that both methods yield the same result and reinforce the properties of quadratic equations and their roots.
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