Let the set of all values of p∈R, for which both the roots of the equation x2−(p+2)x+(2p+9)=0 are negative real numbers, be the interval (α,β]. Then β−2α is equal to:
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Step-by-Step Solution
Step 1: Identify Conditions for Negative Real Roots
For a quadratic equation ax2+bx+c=0 to have two negative real roots, three conditions must be satisfied. First, the discriminant D must be greater than or equal to zero for real roots. Second, the sum of the roots must be negative. Third, the product of the roots must be positive.
Step 2: Apply Discriminant Condition
The discriminant D is calculated as b2−4ac. Substituting the coefficients from the given equation, we get (−(p+2))2−4(1)(2p+9)≥0. Simplifying this inequality leads to p2−4p−32≥0, which factors into (p−8)(p+4)≥0. This inequality holds when p≤−4 or p≥8.
Step 3: Apply Sum of Roots Condition
The sum of the roots is given by -b/a. For negative roots, this sum must be less than zero. Substituting the coefficients, we have −(−(p+2))/1<0, which simplifies to p+2<0. Therefore, p must be less than −2.
Step 4: Apply Product of Roots Condition
The product of the roots is given by c/a. For negative roots, this product must be greater than zero. Substituting the coefficients, we have (2p+9)/1>0, which simplifies to 2p+9>0. This implies 2p>−9, so p>−9/2 or p>−4.5.
Step 5: Find Intersection of Conditions
To find the values of p that satisfy all three conditions, we need to find the intersection of the three intervals. The first condition gives p∈(−∞,−4]∪[8,∞). The second condition gives p∈(−∞,−2). The third condition gives p∈(−4.5,∞). The common interval for all three conditions is (−4.5,−4].
Step 6: Calculate β−2α
From the intersection of the conditions, we found that p belongs to the interval (−4.5,−4]. Comparing this with the given interval (α,β], we have α=−4.5 and β=−4. Now, we can calculate β−2α by substituting these values: −4−2(−4.5)=−4+9=5.