Write the equation \((x-6)^2 = 4x + 5\) in its standard form \(ax^2+bx+c=0\).
Answer and explanation
Correct answer: x^2-16x+31=0
Expand the left side: \((x-6)^2 = x^2 - 12x + 36\). Move the right side \(4x+5\) to the left and combine like terms: \(x^2-12x+36-(4x+5)=x^2-16x+31\). Thus the standard form is \(x^2-16x+31=0\). The closest distractor C (\(x^2-12x+41=0\)) keeps the middle term but has an incorrect constant — a typical arithmetic slip when subtracting the constant. Exam tip: always bring all terms to one side and simplify carefully, checking signs and arithmetic.
Frequently asked questions
What is the correct answer to this question?
x^2-16x+31=0
Why is this the correct answer?
Expand the left side: \((x-6)^2 = x^2 - 12x + 36\). Move the right side \(4x+5\) to the left and combine like terms: \(x^2-12x+36-(4x+5)=x^2-16x+31\). Thus the standard form is \(x^2-16x+31=0\). The closest distractor C (\(x^2-12x+41=0\)) keeps the middle term but has an incorrect constant — a typical arithmetic slip when subtracting the constant. Exam tip: always bring all terms to one side and simplify carefully, checking signs and arithmetic.
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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