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Difficulty: MediumQuadratic Equations
(2x−5)(2x+7)=−k(2x - 5)(2x + 7) = -k

In the given equation, kk is a constant. If the equation has exactly one real solution, what is the value of kk?

Answer: 36

Answer

36
Expanding the equation (2x−5)(2x+7)=−k(2x - 5)(2x + 7) = -k gives 4x2+4x−35=−k4x^2 + 4x - 35 = -k. Adding kk to both sides puts it in standard form 4x2+4x+(k−35)=04x^2 + 4x + (k - 35) = 0. For a quadratic equation to have exactly one real solution, its discriminant must be zero: b2−4ac=42−4(4)(k−35)=0b^2 - 4ac = 4^2 - 4(4)(k - 35) = 0. Simplifying this yields 16−16(k−35)=016 - 16(k - 35) = 0, which gives 1−(k−35)=01 - (k - 35) = 0, so k−35=1k - 35 = 1, and thus k=36k = 36.

Step-by-Step Solution

1
Expand the left side of the equation and write it in standard form.
4x2+4x+(k−35)=04x^2 + 4x + (k - 35) = 0
To apply the quadratic discriminant, the equation must be in the form ax2+bx+c=0ax^2 + bx + c = 0.
2
Identify the coefficients and set the discriminant b2−4acb^2 - 4ac equal to 0.
42−4(4)(k−35)=04^2 - 4(4)(k - 35) = 0
A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.
3
Solve the resulting linear equation for the constant kk.
k=36k = 36
Isolating kk gives the value that satisfies the condition of having exactly one real solution.

Key Concept

Quadratic Discriminant
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Topics

Quadratic Equations

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