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The positive value of k for which x2+kx+64=0 & x2-8x+k=0 will have real roots

The positive value of k for which x2+kx+64=0 & x2-8x+k=0 will have real roots

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The problem states: "The positive value of \( k \) for which \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will have real roots." To find the positive value of \( k \) that ensures both quadratic equations have real roots, start by applying the discriminant condition for real roots. For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( \Delta = b^2 - 4ac \) must be non-negative. 1. **First Equation:** \( x^2 + kx + 64 = 0 \) - Discriminant: \( k^2 - 4 \times 1 \times 64 = k^2 - 256 \) - Condition for real roots: \( k^2 - 256 \geq 0 \) - Simplify: \( k^2 \geq 256 \) - Positive \( k \): \( k \geq 16 \) 2. **Second Equation:** \( x^2 - 8x + k = 0 \) - Discriminant: \( (-8)^2 - 4 \times 1 \times k = 64 - 4k \) - Condition for real roots: \( 64 - 4k \geq 0 \) - Simplify: \( 64 \geq 4k \) - Divide by 4: \( 16 \geq k \) For both conditions to hold, \( k \) must satisfy \( k \geq 16 \) and \( k \leq 16 \). Therefore, the only value that meets both conditions is \( k = 16 \). Thus, the positive value of \( k \) that ensures both equations have real roots is \( k = 16 \).
Transcribed Image Text:The problem states: "The positive value of \( k \) for which \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will have real roots." To find the positive value of \( k \) that ensures both quadratic equations have real roots, start by applying the discriminant condition for real roots. For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( \Delta = b^2 - 4ac \) must be non-negative. 1. **First Equation:** \( x^2 + kx + 64 = 0 \) - Discriminant: \( k^2 - 4 \times 1 \times 64 = k^2 - 256 \) - Condition for real roots: \( k^2 - 256 \geq 0 \) - Simplify: \( k^2 \geq 256 \) - Positive \( k \): \( k \geq 16 \) 2. **Second Equation:** \( x^2 - 8x + k = 0 \) - Discriminant: \( (-8)^2 - 4 \times 1 \times k = 64 - 4k \) - Condition for real roots: \( 64 - 4k \geq 0 \) - Simplify: \( 64 \geq 4k \) - Divide by 4: \( 16 \geq k \) For both conditions to hold, \( k \) must satisfy \( k \geq 16 \) and \( k \leq 16 \). Therefore, the only value that meets both conditions is \( k = 16 \). Thus, the positive value of \( k \) that ensures both equations have real roots is \( k = 16 \).
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