How do we write \(x^2-49=0\) in factorised (factor) form?
Answer and explanation
Correct answer: \((x-7)(x+7)=0\)
\(x^2-49\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\) with \(a=x, b=7\) gives \((x-7)(x+7)\), so the factorised equation is \((x-7)(x+7)=0\). Option B expands to \(x^2-48x-49\), so its middle term is incorrect. Options C and D expand to \(x^2+14x+49\) and \(x^2-14x+49\) respectively, which do not match the original expression. Exam tip: recognize the difference-of-squares pattern quickly and apply \((a-b)(a+b)\).
Frequently asked questions
What is the correct answer to this question?
\((x-7)(x+7)=0\)
Why is this the correct answer?
\(x^2-49\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\) with \(a=x, b=7\) gives \((x-7)(x+7)\), so the factorised equation is \((x-7)(x+7)=0\). Option B expands to \(x^2-48x-49\), so its middle term is incorrect. Options C and D expand to \(x^2+14x+49\) and \(x^2-14x+49\) respectively, which do not match the original expression. Exam tip: recognize the difference-of-squares pattern quickly and apply \((a-b)(a+b)\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
