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Los op vir x: 1.1.1 $x^2 - 7x + 12 = 0$ 1.1.2 $x(3x + 5) = 1$ (korrek tot TWEE desimale syfers) 1.1.3 $x^2 - 2x + 15$ 1.1.4 $ ext{√}(2(1-x)) = x - 1$ Los gelijktidig vir x en y op: 1.2 $3^{x+y} = 27$ en $x^2 + y^2 = 17$ 1.3 Bepaal, sonder die gebruik van 'n sakrekenaar, die waarde van: $ rac{1}{ ext{√}1 + ext{√}2 + ext{√}3 + ext{√}4 + ext{...} + rac{1}{ ext{√}99 + ext{√}100}$ - NSC Mathematics - Question 1 - 2023 - Paper 1

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Question 1

Los-op-vir-x:--1.1.1-$x^2---7x-+-12-=-0$----1.1.2-$x(3x-+-5)-=-1$-(korrek-tot-TWEE-desimale-syfers)----1.1.3-$x^2---2x-+-15$----1.1.4-$-ext{√}(2(1-x))-=-x---1$----Los-gelijktidig-vir-x-en-y-op:--1.2-$3^{x+y}-=-27$-en-$x^2-+-y^2-=-17$--1.3-Bepaal,-sonder-die-gebruik-van-'n-sakrekenaar,-die-waarde-van:--$-rac{1}{-ext{√}1-+--ext{√}2-+--ext{√}3-+--ext{√}4-+--ext{...}-+--rac{1}{-ext{√}99-+--ext{√}100}$-NSC Mathematics-Question 1-2023-Paper 1.png

Los op vir x: 1.1.1 $x^2 - 7x + 12 = 0$ 1.1.2 $x(3x + 5) = 1$ (korrek tot TWEE desimale syfers) 1.1.3 $x^2 - 2x + 15$ 1.1.4 $ ext{√}(2(1-x)) = x - 1$ Lo... show full transcript

Worked Solution & Example Answer:Los op vir x: 1.1.1 $x^2 - 7x + 12 = 0$ 1.1.2 $x(3x + 5) = 1$ (korrek tot TWEE desimale syfers) 1.1.3 $x^2 - 2x + 15$ 1.1.4 $ ext{√}(2(1-x)) = x - 1$ Los gelijktidig vir x en y op: 1.2 $3^{x+y} = 27$ en $x^2 + y^2 = 17$ 1.3 Bepaal, sonder die gebruik van 'n sakrekenaar, die waarde van: $ rac{1}{ ext{√}1 + ext{√}2 + ext{√}3 + ext{√}4 + ext{...} + rac{1}{ ext{√}99 + ext{√}100}$ - NSC Mathematics - Question 1 - 2023 - Paper 1

Step 1

1.1.1 $x^2 - 7x + 12 = 0$

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Answer

To solve for xx, we can factor the quadratic expression:

(x - 3)(x - 4) = 0 ext{Thus, } x = 3 ext{ or } x = 4. \end{align*}$$

Step 2

1.1.2 $x(3x + 5) = 1$ (korrek tot TWEE desimale syfers)

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Answer

Rearranging gives:

3x^2 + 5x - 1 = 0 \end{align*}$$ Using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $a = 3$, $b = 5$, and $c = -1$: $$\begin{align*} b^2 - 4ac & = 5^2 - 4(3)(-1) = 25 + 12 = 37, \end{align*}$$ So: $$x = \frac{-5 \pm \sqrt{37}}{6}$$ Calculating gives: $$x \approx 0.18 ext{ or } x \approx -1.85.$$

Step 3

1.1.3 $x^2 - 2x + 15$

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Answer

Since this is a quadratic that does not factor nicely, we can find the vertex or simply identify that the discriminant (DD) is negative:

D=(2)24(1)(15)=460=56,D = (-2)^2 - 4(1)(15) = 4 - 60 = -56,

This indicates no real roots.

Step 4

1.1.4 $ ext{√}(2(1-x)) = x - 1$

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Answer

Squaring both sides, we obtain:

\\ 2 - 2x = x^2 - 2x + 1$$ Rearranging gives: $$x^2 - 1 = 0 \\ \Rightarrow (x + 1)(x - 1) = 0$$ Thus, $x = 1$ or $x = -1$.

Step 5

1.2 $3^{x+y} = 27$ en $x^2 + y^2 = 17$

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Answer

First, we rewrite 27 as 333^3, leading to:

(y = 3 - x)$$ Substituting into the second equation: $$x^2 + (3-x)^2 = 17$$ This expands and simplifies to: $$2x^2 - 6x + 9 - 17 = 0\ \ \Rightarrow\ 2x^2 - 6x - 8 = 0\ \ \Rightarrow\ x^2 - 3x - 4 = 0$$ Factoring gives: $$(x - 4)(x + 1) = 0,$$ leading to $x = 4$ or $x = -1$. We then find corresponding $y$ values: If $x = 4$, then $y = -1$; If $x = -1$, then $y = 4.$

Step 6

1.3 $ rac{1}{ ext{√}1 + ext{√}2 + ext{√}3 + ext{√}4 + ... + rac{1}{ ext{√}99 + ext{√}100}$

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Answer

Rationalizing the expression breaks this down: The sum reduces to: =1+22+33+4+...+110= -1 + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{4} + ... + \frac{1}{10} \\ Adding terms leads to the conclusion that the overall result sums to 99.

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