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Difficulty: MediumQuadratic Equations and Polynomial Factoring

If 2 is a root of the quadratic equation x2−kx+24=0x^2 - kx + 24 = 0, where kk is a constant, and the quadratic equation x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0 has exactly one real solution, what is the value of mm?

Answer: 64

Answer

The value of mm is 64.
Substituting x=2x = 2 into x2−kx+24=0x^2 - kx + 24 = 0 gives 4−2k+24=04 - 2k + 24 = 0, which simplifies to 2k=282k = 28 or k=14k = 14. Substituting k=14k = 14 into x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0 produces x2−16x+m=0x^2 - 16x + m = 0. For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 to have exactly one real solution, its discriminant b2−4acb^2 - 4ac must equal 0. Therefore, (−16)2−4(1)(m)=0(-16)^2 - 4(1)(m) = 0, which means 256−4m=0256 - 4m = 0, giving m=64m = 64.

Step-by-Step Solution

1
Substitute the known root x=2x = 2 into the equation x2−kx+24=0x^2 - kx + 24 = 0.
k=14k = 14
Since x=2x = 2 is a root of the quadratic equation, evaluating the expression at x=2x = 2 must equal zero.
2
Substitute k=14k = 14 into the second quadratic equation x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0.
x2−16x+m=0x^2 - 16x + m = 0
This determines the linear coefficient of the second quadratic equation.
3
Set the discriminant of x2−16x+m=0x^2 - 16x + m = 0 to zero.
m=64m = 64
A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has exactly one real solution if and only if its discriminant b2−4acb^2 - 4ac equals zero.

Key Concept

Solving quadratic equations via root substitution and applying the discriminant condition for repeated roots.
Estimated Time:1m 30s
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Topics

Quadratic Equations and Polynomial Factoring

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