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If \(x^2-14x+c=0\) is a perfect-square quadratic equation, what is the value of \(c\)?
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Answer and explanation
Correct answer: 49
For the quadratic to be a perfect square, its left-hand polynomial must be expressible as \(x^2-14x+c=(x-7)^2\). Since \((x-7)^2=x^2-14x+49\), we get \(c=49\). Exam tip: for \(x^2+bx+c\) to be a perfect square, \(c=\left(\frac{b}{2}\right)^2\); here, \(b=-14\).
Frequently asked questions
What is the correct answer to this question?
49
Why is this the correct answer?
For the quadratic to be a perfect square, its left-hand polynomial must be expressible as \(x^2-14x+c=(x-7)^2\). Since \((x-7)^2=x^2-14x+49\), we get \(c=49\). Exam tip: for \(x^2+bx+c\) to be a perfect square, \(c=\left(\frac{b}{2}\right)^2\); here, \(b=-14\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
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