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If $1$ is a root of the quadratic equation $3x^2 + ax - 2 = 0$ and the quadratic equation $a(x^2 + 6x) - b = 0$ has equal roots, find the value of b.
Given:
$1$ is a root of the quadratic equation $3x^2 + ax -2 = 0$ and the quadratic equation $a(x^2 + 6x) - b = 0$ has equal roots.
To do:
We have to find the value of b.
Solution:
If $m$ is a root of the quadratic equation $px^2+qx+r=0$ then it satisfies the given equation.
Therefore,
$3x^2 + ax -2 = 0$
$3(1)^2 + a(1) -2 = 0$
$3+a-2=0$
$a+1=0$
$a=-1$
Substituting the value of $a$ in $a(x^2+6x) - b = 0$, we get,
$-1(x^2+6x) - b = 0$
$-x^2-6x-b=0$
$x^2+6x+b=0$
Comparing the quadratic equation $x^2+6x+b=0$ with the standard form of a quadratic equation $px^2+qx+r=0$,
$p=1, q=6$ and $r=b$
The discriminant of the standard form of the quadratic equation $px^2+qx+r=0$ is $D=q^2-4pr$.
$D=(6)^2-4(1)(b)$
$D=36-4b$
The given quadratic equation has equal roots if $D=0$.
Therefore,
$36-4b=0$
$36=4b$
$b=\frac{36}{4}$
$b=9$
The value of $b$ is $9$.
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