Question
Show that the equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ has no real roots, when $\text{a}\neq\text{b}.$
✓
Answer
The quadric equation is $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$
Here, $a = 2(a^2 + b^2), b = 2(a + b)$ and $c = 1$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 2(a^2 + b^2), b = 2(a + b)$ and c$ = 1$
$\Rightarrow D = {2(a + b)}^2 - 4 \times 2(a^2 + b^2) \times 1$
$\Rightarrow D = 4(a^2 + 2ab + b^2) - 8(a^2 + b^2) \times 1$
$\Rightarrow D = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2$
$\Rightarrow D = 8ab - 4a^2 - 4b^2$
$\Rightarrow D = -4(a^2 - 2ab + b^2)$
$\Rightarrow D = -4(a - b)^2$
We have, $\text{a}\neq\text{b}$
$\Rightarrow\text{a}-\text{b}\neq0$
Thus, the value of $D < 0$
Therefore, the roots of the given equation are not real.
Hence, proved.
Here, $a = 2(a^2 + b^2), b = 2(a + b)$ and $c = 1$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 2(a^2 + b^2), b = 2(a + b)$ and c$ = 1$
$\Rightarrow D = {2(a + b)}^2 - 4 \times 2(a^2 + b^2) \times 1$
$\Rightarrow D = 4(a^2 + 2ab + b^2) - 8(a^2 + b^2) \times 1$
$\Rightarrow D = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2$
$\Rightarrow D = 8ab - 4a^2 - 4b^2$
$\Rightarrow D = -4(a^2 - 2ab + b^2)$
$\Rightarrow D = -4(a - b)^2$
We have, $\text{a}\neq\text{b}$
$\Rightarrow\text{a}-\text{b}\neq0$
Thus, the value of $D < 0$
Therefore, the roots of the given equation are not real.
Hence, proved.
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