Question
If the roots the equations $ax^2 + 2bx + c = 0$ and $\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0$ are simultaneously real, then prove that $b^2 = ac.$
✓
Answer
The given equation are
$\text{ax}^2+2\text{bx}+\text{c}=0\ ....(\text{i})$
$\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0\ .....(\text{ii})$
Roots are simultaneously real,
Then prove that $\text{b}^2=\text{ac}$
Let $D_1$ and $D_2$ be the discriminants of equation (i) and (ii) respectively.
Then, $\text{D}_1=(2\text{b})^2-4\text{ac}$
$\text{D}_1=4\text{b}^2-4\text{ac}$
And, $\text{D}_2=(-2\sqrt{\text{ac}})^2-4\times\text{b}\times\text{b}$
$\text{D}_2=4\text{ac}-4\text{b}^2$
Both the given equation will have real roots, if $\text{D}_1\geq0$ and $\text{D}_2\geq0$
$4\text{b}^2-4\text{ac}\geq0$
$4\text{ac}\geq4\text{b}^2$
$\text{b}^2\geq\text{ac}\ ....(\text{iii})$
$4\text{ac}-4\text{b}^2\geq0$
$4\text{ac}\geq4\text{b}^2$
$\text{ac}\geq\text{b}^2\ ....(\text{iv})$
From equations (iii) and (iv) we get
$\text{b}^2=\text{ac}$
Hence, $\text{b}^2=\text{ac}$
$\text{ax}^2+2\text{bx}+\text{c}=0\ ....(\text{i})$
$\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0\ .....(\text{ii})$
Roots are simultaneously real,
Then prove that $\text{b}^2=\text{ac}$
Let $D_1$ and $D_2$ be the discriminants of equation (i) and (ii) respectively.
Then, $\text{D}_1=(2\text{b})^2-4\text{ac}$
$\text{D}_1=4\text{b}^2-4\text{ac}$
And, $\text{D}_2=(-2\sqrt{\text{ac}})^2-4\times\text{b}\times\text{b}$
$\text{D}_2=4\text{ac}-4\text{b}^2$
Both the given equation will have real roots, if $\text{D}_1\geq0$ and $\text{D}_2\geq0$
$4\text{b}^2-4\text{ac}\geq0$
$4\text{ac}\geq4\text{b}^2$
$\text{b}^2\geq\text{ac}\ ....(\text{iii})$
$4\text{ac}-4\text{b}^2\geq0$
$4\text{ac}\geq4\text{b}^2$
$\text{ac}\geq\text{b}^2\ ....(\text{iv})$
From equations (iii) and (iv) we get
$\text{b}^2=\text{ac}$
Hence, $\text{b}^2=\text{ac}$
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