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12 Mar 2018, 05:51
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What is the largest integral value of 'k' for which the quadratic equation x2 - 6x + k = 0 will have two real and distinct roots?
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Using discriminant:
\(x1,2=\frac{6+\sqrt{36-4k}}{2}\)=\(\frac{6+\sqrt{4*(9-k)}}{2}\)=\(\frac{6+2\sqrt{(9-k)}}{2}\)=\({3+\sqrt{(9-k)}}\). So, for the expression to have distinct real roots, \(\sqrt{9-k}\) has to be an integer, therefore k can:
k=5, x1=\({3+\sqrt{(9-5)}}=3+2=5\), or x2=\({3+\sqrt{(9-5)}}=3-2=1\)
k=8, x1=\({3+\sqrt{(9-8)}}=3+1=4\), or x2=\({3+\sqrt{(9-5)}}=3-1=2\)
k=9, x=\({3+\sqrt{(9-9)}}=3\)
So, the expression becomes \(x^2-6x+9=(x-3)^2=0\)
\(x1,2=\frac{6+\sqrt{36-4k}}{2}\)=\(\frac{6+\sqrt{4*(9-k)}}{2}\)=\(\frac{6+2\sqrt{(9-k)}}{2}\)=\({3+\sqrt{(9-k)}}\). So, for the expression to have distinct real roots, \(\sqrt{9-k}\) has to be an integer, therefore k can:
k=5, x1=\({3+\sqrt{(9-5)}}=3+2=5\), or x2=\({3+\sqrt{(9-5)}}=3-2=1\)
k=8, x1=\({3+\sqrt{(9-8)}}=3+1=4\), or x2=\({3+\sqrt{(9-5)}}=3-1=2\)
k=9, x=\({3+\sqrt{(9-9)}}=3\)
So, the expression becomes \(x^2-6x+9=(x-3)^2=0\)
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13 Mar 2018, 14:51
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I've made a mistake in my previous post. The answer is 9 using Vieta's formula:
\(x^2-6x+k=0\)
x1+x2=6
x1*x2=k
x1 and x2 must be > 0 to maximize k and x1 and x2 cannot have different signs:
x1=1, x2=5 => x1*x2=k=5
x1=2, x2=4 => x1*x2=k=8
x1=3, x2=3 => x1*x2=k=9
\(x^2-6x+k=0\)
x1+x2=6
x1*x2=k
x1 and x2 must be > 0 to maximize k and x1 and x2 cannot have different signs:
x1=1, x2=5 => x1*x2=k=5
x1=2, x2=4 => x1*x2=k=8
x1=3, x2=3 => x1*x2=k=9
