2 Marks Questions

Question 512 Marks

Solve: $\text{3x}^2+5\sqrt5\text{x}-10=0$

Answer

$\text{3x}^2+5\sqrt5\text{x}-10=0$$\Rightarrow\text{3x}^2+6\sqrt5\text{x}-\sqrt5\text{x}-10=0$
$\Rightarrow\text{3x}\big(\text{x}+2\sqrt5\big)-\sqrt5\big(\text{x}+2\sqrt5\big)=0$
$\Rightarrow\big(\text{x}+2\sqrt5\big)\big(\text{3x}-\sqrt5\big)=0$
$\Rightarrow\text{x}+2\sqrt5=0$ or $\text{3x}-\sqrt5=0$
$\Rightarrow\text{x}=-2\sqrt5$ or $\text{x}=\frac{\sqrt5}{3}$

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Question 522 Marks

Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 4x - 1 = 0$

Answer

Given, $x^2 - 4x - 1 = 0$ On comparing it with $ax^2 + bx + c = 0,$
we get : $a = 1, b = -4$ and $c = -1$ Discriminant $D$ is
given by : $D = (b^2 - 4ac) = (-4)^2 - 4 \times 1 \times (-1) = 16 + 4 = 20 = 20 > 0$
Hence, the roots of the equation are real. Roots $\alpha$ and $\beta$ are given by: $\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-4)+\sqrt{20}}{2\times1}$
$=\frac{4+2\sqrt5}{2}$
$=\frac{2\big(2+\sqrt5\big)}{2}$
$=\big(2+\sqrt5\big)$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-4)-\sqrt{20}}{2}$
$=\frac{4-2\sqrt5}{2}$
$=\frac{2\big(2-\sqrt5\big)}{2} $
$=\big(2-\sqrt5\big)$
Thus, the roots of the equation are $\big(2+\sqrt5\big)$ and $\big(2-\sqrt5\big)$

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Question 532 Marks

Solve: $\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$

Answer

The given equation is $\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=1,\ \text{b}=-\big(\sqrt3+1\big),\ \text{c}=\sqrt3$
$\therefore\text{D}=\text{b}^2-\text{4ac}$
$=\big[-\big(\sqrt3+1\big)\big]^2-4\times1\times\sqrt3$
$=\big(\sqrt3-1\big)^2>0$
So, the given equation has real roots, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]+\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1+\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]-\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1-\sqrt3-1}{2}$
$=\frac{2}{2}$
$=1$
Hence, $\sqrt3$ and 1 are the roots of the given equation.

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Question 542 Marks

Solve the following quadratic equation:$x^2 = 18x - 77$

Answer

$x^2 = 18x - 77$
$\Rightarrow x^2 - 18x + 77 = 0$
$\Rightarrow x^2 - 11x - 7x + 77 = 0$
$\Rightarrow x(x - 11) - 7(x - 11) = 0$
$\Rightarrow (x - 11)(x - 7) = 0$
$\Rightarrow x - 11 = 0 or x - 7 = 0$
$\Rightarrow x = 11 or x = 7$

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Question 552 Marks

Solve: $\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$

Answer

$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2+\text{12x}-\text{2x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\big(\text{x}+4\sqrt3\big)\big(\sqrt3\text{x}-2\big)=0$
$\Rightarrow\text{x}+4\sqrt3=0$ or $\sqrt3\text{x}-2=0$
$\Rightarrow\text{x}=-4\sqrt3$ or $\text{x}=\frac{2}{\sqrt3}$

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Question 562 Marks

Find the discriminant of the following equation:
$1 - x = 2x^2$

Answer

$1 - x = 2x^2$
$2x^2+ x - 1 = 0$
Here,
$a = 2,$
$b = 1,$
$c = -1$
Discriminant D is given by:
$D = b^2 - 4ac$
$= 1^2 - 4 \times 2(-1)$
$= 1 + 8$
$= 9$

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Question 572 Marks

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 $\text{x}=-\frac{1}3{}$

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Question 582 Marks

Find the nature of the roots of the following quadratic equations:$5x(x - 2) + 6 = 0$

Answer

The given equation is $5x(x - 2) + 6 = 0$
$\Rightarrow 5x^2 - 10x + 6 = 0$
This is of the form $ax^2 + bx + c = 0$, where $a = 5, b = -10$ and $c = 6$
$\therefore$ Discriminant, $D = b^2- 4ac$
$D = (-10)^2 - 4 \times 5 \times 6$
$D = 100 - 120$
$D = -20 > 0$
Hence, the given equation has no real roots.

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Question 592 Marks

Solve the following quadratic equation:$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$

Answer

$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$ Here, $6\sqrt3\times\sqrt3=6\times3=18$ and 9 × 2 = 18 & 9 + 2 = 11 $\Rightarrow\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$$\Rightarrow\sqrt3\text{x}^2+9\text{x}+2\text{x}+6\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+3\sqrt3\big)+2\big(\text{x}+3\sqrt{3}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{3}\big)\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\text{x}+3\sqrt3=0$ or $\sqrt3\text{x}+2=0$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2}{\sqrt3}$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2\times\sqrt3}{\sqrt3\times\sqrt3}=\frac{-2\sqrt3}{3}$
Hence, $-3\sqrt3$ and $\frac{-2\sqrt3}{3}$ are the roots of the given equation.

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Question 602 Marks

Solve the following quadratic equation:$6x^2 + 11x + 3 = 0$

Answer

$6x^2 + 11x + 3 = 0$
$\Rightarrow 6x^2 + 9x + 2x + 3 = 0$
$\Rightarrow 3x(2x + 3) + 1(2x + 3) = 0$
$\Rightarrow (2x + 3)(3x + 1) = 0$
$\Rightarrow (2x + 3) = 0 or (3x + 1) = 0$
$\Rightarrow\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{-1}{3}$
Hence, $\frac{-3}{2}$ and $\frac{-1}{3}$ are the roots of $6x^2 + 11x + 3 = 0$

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Maths 2 Marks Questions