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Standard XI

Mathematics

Algebra of Roots of Quadratic Equations

If x^2+px + q...

Question

If $x^2+px + q = 0$ is the quadratic equation whose roots are
$a~– 2$ and $b~– 2$ where $a$ and $b$ are the roots of $x^2 - 3x + 1 = 0$, then

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Solution

$a + b = 3, ab = 1$ and $–p = a~– 2 + b~– 2, q = (a~– 2)(b~– 2)$

$\Rightarrow~–p = a + b~– 4, q = ab~– 2(a + b) + 4$

$\Rightarrow$ $–p = 3~–~4$ and $q = 1~– ~2(3) + 4$
$\Rightarrow (p, q) = (1, –1)$

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If \(x^2+px + q = 0\) is the quadratic equation whose roots are
\(a~– 2\) and \(b~– 2\) where \(a\) and \(b\) are the roots of \(x^2 - 3x + 1 = 0\), then

\(a + b = 3, ab = 1\) and \(–p = a~– 2 + b~– 2, q = (a~– 2)(b~– 2)\)

\(\Rightarrow~–p = a + b~– 4, q = ab~– 2(a + b) + 4\)

\(\Rightarrow\) \(–p = 3~–~4\) and \(q = 1~– ~2(3) + 4\)
\(\Rightarrow (p, q) = (1, –1)\)

If \(x^2+px + q = 0\) is the quadratic equation whose roots are \(a~– 2\) and \(b~– 2\) where \(a\) and \(b\) are the roots of \(x^2 - 3x + 1 = 0\), then

\(a + b = 3, ab = 1\) and \(–p = a~– 2 + b~– 2, q = (a~– 2)(b~– 2)\) \(\Rightarrow~–p = a + b~– 4, q = ab~– 2(a + b) + 4\) \(\Rightarrow\) \(–p = 3~–~4\) and \(q = 1~– ~2(3) + 4\) \(\Rightarrow (p, q) = (1, –1)\)

If \(x^2+px + q = 0\) is the quadratic equation whose roots are
\(a~– 2\) and \(b~– 2\) where \(a\) and \(b\) are the roots of \(x^2 - 3x + 1 = 0\), then

\(a + b = 3, ab = 1\) and \(–p = a~– 2 + b~– 2, q = (a~– 2)(b~– 2)\)

\(\Rightarrow~–p = a + b~– 4, q = ab~– 2(a + b) + 4\)

\(\Rightarrow\) \(–p = 3~–~4\) and \(q = 1~– ~2(3) + 4\)
\(\Rightarrow (p, q) = (1, –1)\)