From x2 − 5x + 6 = 0, find the roots using Vieta.
Solution
Correct Answer: Option A
Correct answer: Option 1 (2 and 3). Explanation (textbook style, using Vieta and factoring):
For the quadratic \(x^{2}-5x+6=0\) (here \(a=1,\;b=-5,\;c=6\)), Vieta's formulas state that if the roots are \(r_1,r_2\) then
\(\;r_1+r_2=-\dfrac{b}{a}\) and \(\;r_1r_2=\dfrac{c}{a}\).
Thus \(r_1+r_2=-\dfrac{-5}{1}=5\) and \(r_1r_2=\dfrac{6}{1}=6\). We need two numbers whose sum is \(5\) and product is \(6\); these numbers are \(2\) and \(3\). Therefore the roots are \(2\) and \(3\).
Equivalently, factor the quadratic: \(x^{2}-5x+6=(x-2)(x-3)\). Setting this to zero gives \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\).
For the quadratic \(x^{2}-5x+6=0\) (here \(a=1,\;b=-5,\;c=6\)), Vieta's formulas state that if the roots are \(r_1,r_2\) then
\(\;r_1+r_2=-\dfrac{b}{a}\) and \(\;r_1r_2=\dfrac{c}{a}\).
Thus \(r_1+r_2=-\dfrac{-5}{1}=5\) and \(r_1r_2=\dfrac{6}{1}=6\). We need two numbers whose sum is \(5\) and product is \(6\); these numbers are \(2\) and \(3\). Therefore the roots are \(2\) and \(3\).
Equivalently, factor the quadratic: \(x^{2}-5x+6=(x-2)(x-3)\). Setting this to zero gives \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\).
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