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The quadratic equation whose roots are 2 + √3 and 2 - √3 is:
x² - 4x + 1 = 0
x² + 4x + 1 = 0
x² - 4x - 1 = 0
x² + 4x - 1 = 0
x² - 4x + 1 = 0
Quickly find the sum of roots (S=2+3+2−3=4) and product of roots (P=(2+3)(2−3)=4−3=1), then substitute into x2−Sx+P=0.
Roots of the quadratic equation are α=2+3and β=2−3
x2−(S)x+P=0
Quickly find the sum of roots (S=2+3+2−3=4) and product of roots (P=(2+3)(2−3)=4−3=1), then substitute into x2−Sx+P=0.
Forgetting that (a+b)(a−b)=a2−b, which leads to calculation errors in the product of roots.
Identify the given roots
Let the roots of the quadratic equation be given as α=2+3 and β=2−3.
α=2+3,β=2−3
Calculate the sum of the roots
Add the two roots together: S=α+β=(2+3)+(2−3)=4.
S=(2+3)+(2−3)=4
Calculate the product of the roots
Multiply the two roots using the difference of squares formula: P=α⋅β=(2+3)(2−3)=22−(3)2=4−3=1.
P=(2+3)(2−3)=4−3=1
Form the quadratic equation
Substitute the sum and product into the standard quadratic equation formula: x2−Sx+P=0, which gives x2−4x+1=0.
x2−4x+1=0
A is correct because the sum of the roots is 4 and the product is 1, yielding the equation x2−4x+1=0.
Understanding the relationship between roots and coefficients is crucial for higher-level polynomial equations and complex number problems in competitive exams.
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