Question
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$\big(\text{x}-\sqrt{2}\big)^2-2(\text{x}+1)=0.$
$\big(\text{x}-\sqrt{2}\big)^2-2(\text{x}+1)=0.$
✓
Answer
Main concept used:
Quadratic equation $a x^2+b x+c=0$ will have two distinct real roots if $D >0$ or $b ^2-4 ac >0$.
$(x-\sqrt{2})^2-2(x+1)=0$
$\Rightarrow x^2-2 \sqrt{2} x+2-2 x-2=0$
$\Rightarrow x^2-(2 \sqrt{2}+2) x=0$
Now, $D=b^2-4 a c$
$=[-(2 \sqrt{2}+2)]^2-4 \times 1 \times 0$
${[\therefore a=1, b=-(2 \sqrt{2}+2), c=0]}$
$=(2 \sqrt{2}+2)^2>0$
As $D>0$, so the given equation has real and unequal roots.
Quadratic equation $a x^2+b x+c=0$ will have two distinct real roots if $D >0$ or $b ^2-4 ac >0$.
$(x-\sqrt{2})^2-2(x+1)=0$
$\Rightarrow x^2-2 \sqrt{2} x+2-2 x-2=0$
$\Rightarrow x^2-(2 \sqrt{2}+2) x=0$
Now, $D=b^2-4 a c$
$=[-(2 \sqrt{2}+2)]^2-4 \times 1 \times 0$
${[\therefore a=1, b=-(2 \sqrt{2}+2), c=0]}$
$=(2 \sqrt{2}+2)^2>0$
As $D>0$, so the given equation has real and unequal roots.
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