If the roots of the equations $ax^2+2bx+c=0$ and $bx^2-2\sqrt{ac}x+b=0$ are simultaneously real, then prove that $b^2=ac$.
Given:
Given quadratic equations are $ax^2+2bx+c=0$ and $bx^2-2\sqrt{ac}x+b=0$. The roots of the given quadratic equations are simultaneously real.
To do:
We have to prove that $b^2=ac$.
Solution:
Let $D_1$ be the discriminant of $ax^2+2bx+c=0$ and $D_2$ be the discriminant of $bx^2-2\sqrt{ac}x+b=0$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
Therefore,
$D_1=(2b)^2-4(a)(c)$
$D_1=4b^2-4ac$
$D_2=(-2\sqrt{ac})^2-4(b)(b)$
$D_2=4ac-4b^2$
The given quadratic equations have real roots if $D_1≥0$ and $D_2≥0$.
This implies,
$4b^2-4ac≥0$ and $4ac-4b^2≥0$
$b^2-ac≥0$ and $ac-b^2≥0$
$b^2≥ac$ and $ac≥b^2$
Therefore,
$b^2=ac$
Hence proved.
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