If \(x=4\) is a root of the equation \(x^2+mx-20=0\), what is the value of \(m\)?
Answer and explanation
Correct answer: 1
Substitute \(x=4\) into the equation: \(16+4m-20=0\), so \(4m-4=0\) and hence \(m=1\). The nearest distractor (\(-1\)) is wrong because with \(m=-1\) the left side becomes \(16-4-20=-8\), not zero. Exam tip: you can also use Vieta — with one root 4 the other is \(-5\) (product = \(-20\)), sum = \(4+(-5)=-1=-m\) giving \(m=1\).
Frequently asked questions
What is the correct answer to this question?
1
Why is this the correct answer?
Substitute \(x=4\) into the equation: \(16+4m-20=0\), so \(4m-4=0\) and hence \(m=1\). The nearest distractor (\(-1\)) is wrong because with \(m=-1\) the left side becomes \(16-4-20=-8\), not zero. Exam tip: you can also use Vieta — with one root 4 the other is \(-5\) (product = \(-20\)), sum = \(4+(-5)=-1=-m\) giving \(m=1\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
