Which is the correct factorised form of the quadratic equation \(2x^2-7x+5=0\)?
Answer and explanation
Correct answer: \((2x-5)(x-1)=0\)
Split the middle term as \(-2x-5x\): \(2x^2-7x+5=2x^2-2x-5x+5=(2x-5)(x-1)\). Hence, the factorised equation is \((2x-5)(x-1)=0\). Option C is incorrect because its expansion gives the middle term \(-11x\), not \(-7x\). As an exam tip, expand the factors once to verify the middle term and constant term.
Frequently asked questions
What is the correct answer to this question?
\((2x-5)(x-1)=0\)
Why is this the correct answer?
Split the middle term as \(-2x-5x\): \(2x^2-7x+5=2x^2-2x-5x+5=(2x-5)(x-1)\). Hence, the factorised equation is \((2x-5)(x-1)=0\). Option C is incorrect because its expansion gives the middle term \(-11x\), not \(-7x\). As an exam tip, expand the factors once to verify the middle term and constant term.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Methods of Solving Quadratic Equations.
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