Question
Show that the equation $2\left(a^2+b^2\right) x^2+2(a+b) x+1=0$ has no real roots, when $\text{a}\neq\text{b}.$
✓
Answer
The quadric equation is $2\left(a^2+b^2\right) x^2+2(a+b) x+1=0$
Here, $a=2\left(a^2+b^2\right), b=2(a+b)$ and $c=1$
As we know that $D=b^2-4 a c$
Putting the value of $a=2\left(a^2+b^2\right), b=2(a+b)$ and $c=1$
$ \Rightarrow D=\{2(a+b)\}^2-4 \times 2\left(a^2+b^2\right) \times 1 $
$ \Rightarrow D=4\left(a^2+2 a b+b^2\right)-8\left(a^2+b^2\right) \times 1 $
$ \Rightarrow D=4 a^2+8 a b+4 b^2-8 a^2-8 b^2 $
$ \Rightarrow D=8 a b-4 a^2-4 b^2 $
$ \Rightarrow D=-4\left(a^2-2 a b+b^2\right) $
$ \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, $a=2\left(a^2+b^2\right), b=2(a+b)$ and $c=1$
As we know that $D=b^2-4 a c$
Putting the value of $a=2\left(a^2+b^2\right), b=2(a+b)$ and $c=1$
$ \Rightarrow D=\{2(a+b)\}^2-4 \times 2\left(a^2+b^2\right) \times 1 $
$ \Rightarrow D=4\left(a^2+2 a b+b^2\right)-8\left(a^2+b^2\right) \times 1 $
$ \Rightarrow D=4 a^2+8 a b+4 b^2-8 a^2-8 b^2 $
$ \Rightarrow D=8 a b-4 a^2-4 b^2 $
$ \Rightarrow D=-4\left(a^2-2 a b+b^2\right) $
$ \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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