The number of integral values of \(k\), for which one root of the equation \[2x^2 - 8x + k = 0\] lies in the interval \((1, 2)\) and its other root lies in the interval \((2, 3)\), is:
For problems involving roots of quadratic equations in specific intervals, use the condition f(a) · f(b) < 0 to identify intervals of k and solve the resulting inequalities systematically.
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The given quadratic equation is:
\[2x^2 - 8x + k = 0.\]
Step 1: Conditions for Roots in the Specified Intervals
For one root to lie in \((1, 2)\), the product of the values of the quadratic function at the endpoints must be negative:
\[f(1) \cdot f(2) < 0.\]
Similarly, for the other root to lie in \((2, 3)\):
\[f(2) \cdot f(3) < 0.\]
Step 2: Compute \(f(x)\) at Specific Points
The quadratic function is:
\[f(x) = 2x^2 - 8x + k.\]
Compute \(f(1)\), \(f(2)\), and \(f(3)\):
\[f(1) = 2(1)^2 - 8(1) + k = 2 - 8 + k = k - 6,\]
\[f(2) = 2(2)^2 - 8(2) + k = 8 - 16 + k = k - 8,\]
\[f(3) = 2(3)^2 - 8(3) + k = 18 - 24 + k = k - 6.\]
Step 3: Apply the Conditions
For \(f(1) \cdot f(2) < 0\):
\[(k - 6)(k - 8) < 0.\]
The roots of this inequality are \(k = 6\) and \(k = 8\). Using the sign change rule for quadratic inequalities:
\[k \in (6, 8).\]
For \(f(2) \cdot f(3) < 0\):
\[(k - 8)(k - 6) < 0.\]
This inequality is also satisfied for:
\[k \in (6, 8).\]
Step 4: Determine Integral Values of \(k\)
The intersection of both conditions is \(k \in (6, 8)\). The only integer in this interval is:
\[k = 7.\]
Conclusion: The number of integral values of \(k\) is \(\mathbf{1}\) (Option 3).
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
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