Question
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$(x + 4)^2 - 8x = 0.$
$(x + 4)^2 - 8x = 0.$
✓
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$(x + 4)^2 - 8x = 0$
$\Rightarrow (x)^2 + (4)^2 + 2(x)(4) - 8x = 0$
$\Rightarrow x^2 + 16 + 8x - 8x = 0$
$\Rightarrow x^2 + 16 = 0$
$\Rightarrow x^2 + 0x + 16 = 0$
Now, $D = b^2 - 4ac$
$= (0)^2 - 4(1) (16) (a = 1, b = 0, c = 16)$
$\Rightarrow D = -64 < 0$
$As D < 0$, so the given equation has no real roots.
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$(x + 4)^2 - 8x = 0$
$\Rightarrow (x)^2 + (4)^2 + 2(x)(4) - 8x = 0$
$\Rightarrow x^2 + 16 + 8x - 8x = 0$
$\Rightarrow x^2 + 16 = 0$
$\Rightarrow x^2 + 0x + 16 = 0$
Now, $D = b^2 - 4ac$
$= (0)^2 - 4(1) (16) (a = 1, b = 0, c = 16)$
$\Rightarrow D = -64 < 0$
$As D < 0$, so the given equation has no real roots.
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