Question
Solve the following quadratic equation:$9x^2 - 3x - 2 = 0$
✓
Answer
$9x^2 - 3x - 2 = 0$
$\Rightarrow 9x^2 - 6x + 3x - 2 = 0$
$\Rightarrow 3x(3x - 2) + 1(3x - 2) = 0$
$\Rightarrow (3x - 2)(3x + 1) = 0$
$\Rightarrow 3x - 2 = 0 or 3x + 1 = 0$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}3{}$
$\Rightarrow 9x^2 - 6x + 3x - 2 = 0$
$\Rightarrow 3x(3x - 2) + 1(3x - 2) = 0$
$\Rightarrow (3x - 2)(3x + 1) = 0$
$\Rightarrow 3x - 2 = 0 or 3x + 1 = 0$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}3{}$
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In figure 3.98, seg AB is a diameter of a circle with centre O . The bisector of∠ ACB intersects the circle at point D. Prove that, seg AD ≅seg BD.
Complete the following proof by filling in the blanks.
Proof: Draw seg OD.
$\angle ACB =\square \ldots 90^{\circ} \ldots \ldots \ldots . . \text { angle inscribed in semicircle }$
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→Complete the following proof by filling in the blanks.
Proof: Draw seg OD.
$\angle ACB =\square \ldots 90^{\circ} \ldots \ldots \ldots . . \text { angle inscribed in semicircle }$
$\angle DCB =\square \ldots . . .45^{\circ} \ldots CD \text { is the bisector of } \angle C$
$m (\operatorname{arc} DB )=\square . . . .45^{\circ} \ldots \text { inscribed angle theorem }$
$\angle DOB =\square \ldots . . . .90^{\circ} \text { definition of measure of an arc (I) }$
$\operatorname{seg} OA \cong \operatorname{seg} OB \ldots \ldots . . \text { radii of the circle... } \square \text { (II) }$
$\therefore \text { line } OD \text { is } \square \text { of seg } AB . \ldots . . \text { bisector.... From (I) and (II) }$
$\therefore \text { seg } AD \cong \operatorname{seg} BD$
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