Maths M.C.Q (1 Marks)

$7x^2- 12x + 18 = 0$
Comparing with $ax^2+ bx + c = 0$, we have
$a = 7, b = -12, c = 18$
Sum of the roots $=-\frac{\text{b}}{\text{a}}=-\frac{(-12)}{7}=\frac{12}{7}$
Product of the roots $=\frac{\text{c}}{\text{a}}=\frac{18}{7}$
Now,
$\frac{\text{Sum of the roots}}{\text{Product of the roots}}=\frac{\frac{12}{7}}{\frac{18}{7}}=\frac{12}{18}$
$=\frac{2}{3}=2:3$

Since $x = 3$ is a solution of the equation $3x^2 + (k - 1)x + 9 = 0$, we have
$3(3)^2 + (k - 1)3 + 9 = 0$
$\Rightarrow 27 + 3k - 3 + 9 = 0$
$\Rightarrow 3k + 33 = 0$
$\Rightarrow 3k = -33$
$\Rightarrow k = -11$

Since the equation $x^2+5 k x+16=0$ has no real roots,
$\Rightarrow D<0$
$\Rightarrow b^2-4 a c>0$
$\Rightarrow(5 k)^2-4 \times 16<0$
$\Rightarrow 25 k^2-64<0$
$\Rightarrow 25 k^2<64$
$\Rightarrow k^2<\frac{64}{25}$
$\Rightarrow k<\sqrt{\frac{64}{25}} \text { or } k>-\sqrt{\frac{64}{25}}$
$\Rightarrow k<\frac{8}{5} \text { or } k>-\frac{8}{5}$
$\Rightarrow-\frac{8}{5}

Let one root of the given equation be $\alpha$.
Then, its root will be $\frac{1}{\alpha}$.
Given equation is $5 x^2+13 x+k=0$
Comparing with $ax ^2+ bx + c =0$, we have
$a=5, b=13, c=k$
Now,
Product of the roots $=\frac{ c }{ a }$
$\Rightarrow \alpha \times \frac{1}{\alpha}=\frac{k}{5}$
$\Rightarrow k=5$

Given equation is $kx^2 + 2x + 3x = 0$
Comparing with $ax^2+ bx + c = 0$, we have
$a = k, b = 2, c = 3k$
Now,
Sum of the roots $=$ product of the roots
$\Rightarrow\frac{-2}{\text{k}}=\frac{\text{3k}}{\text{k}}$
$\Rightarrow\text{3k}=-2$
$\Rightarrow\text{k}=\frac{-2}3{}$

Given equation $2 x^2-6 x+3=0$
Here, $a=2, b=-6, c=3$
Discriminant, $D=b^2-4 a c$
$=(-6)^2-4 \times 2 \times 3$
$=36-24$
$=12<0$
Also, $12$ is not a perfect square.
Hence, the roots of the given equation are real, unequal and irrational.

$x^2 - 6x + 2 = 0$
Comparing with $ax^2+ bx + c = 0$, we have
$a = 1, b = -6, c = 2$
Sum of the roots $=-\frac{\text{b}}{\text{a}}=-\frac{(-6)}{1}=6$

Given, the roots of $kx^2 - 6x - 2 = 0$ are real
$\Rightarrow\text{D}\ge0$
$\Rightarrow\text{b}^2-\text{4ac}\ge0$
$\Rightarrow(-6)^2-4\times\text{k}\times(-2)\ge0$
$\Rightarrow36+\text{8k}\ge0$
$\Rightarrow\text{8k}\ge-36$
$\Rightarrow\text{k}\ge-\frac{9}{2}$

Let the other root be $\alpha.$
Given equation is $3x^2- 10x + 3 = 0$
Comparing with $ax^2+ bx + c = 0$, we have
$a = 3, b = -10, c = 3$
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\times\frac{1}{3}=\frac{3}{3}$
$\Rightarrow\alpha=3$

Let the breadth of the rectangle be $x m$.
Then, length of the rectangle $=(x+8) m$
$\text { Now Area }=240 m^2$
$\Rightarrow \text { Length } \times \text { Breadth }=240$
$\Rightarrow x(x+8)=240$
$\Rightarrow x^2+8 x=240$
$\Rightarrow x^2+8 x-240=0$
$\Rightarrow x^2+20 x-12 x-240=0$
$\Rightarrow x(x+20)-12(x+20)=0$
$\Rightarrow(x+20)(x-12)=0$
$\Rightarrow x+20=0 \text { or } x-12=0$
$\Rightarrow x=-20 \text { or } x=12$
$\Rightarrow x=12 \text { (Breadth cannot be negative). }$

1. $(x^2 + 1) = (2 - x)^2 + 3$

2. $x^3 - x^2= (x - 1)^2$

3. $2x^2 + 3 = (5 + x)(2x - 3)$

Product of the roots $=\frac{\text{c}}{\text{a}}$
Also, ${\alpha}\times\frac{1}{\alpha}=1$
$\Rightarrow\frac{\text{c}}{\text{a}}=1$
$\Rightarrow\text{c}=\text{a}$

Let the number be $x.$
Then, $\text{x}+\frac{1}{\text{x}}=\frac{41}{20}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{41}{20}$
$\Rightarrow 20x^2 + 20 = 41x$
$\Rightarrow 20x^2 - 41x + 20 = 0$
$\Rightarrow 20x^2 - 25x - 16x + 20 = 0$
$\Rightarrow 5x(4x - 5) - 4(4x - 5) = 0$
$\Rightarrow (4x - 5)(5x - 4) = 0$
$\Rightarrow 4x - 5 = 0 or 5x - 4 = 0$
$\Rightarrow 4x = 5 or 5x = 4$
$\Rightarrow\text{x}=\frac{5}4{}$ or $\text{x}=\frac{4}{5}$

Since the roots of the equation $4x^2- 3kx + 1 = 0$ are equal,
$D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow (-3k)^2 - 4 \times 4 \times 1 = 0$
$\Rightarrow 9k^2 - 16 = 0$
$\Rightarrow 9k^2 = 16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\Rightarrow\text{k}=\pm\frac{4}{3}$

Since the roots of the equation $ax^2 + bx + c = 0,$ $\text{a}\neq0$ are real and unequal,
we must have $D > 0$
$\Rightarrow b^2 - 4ac > 0$

Given, the roots of $5x^2 - kx + 1 = 0$ are real and distinct.
$\Rightarrow D > 0$
$\Rightarrow b^2 - 4ac > 0$
$\Rightarrow (-k)^2 - 4 \times 5 \times 1 > 0$
$\Rightarrow k^2 - 20 > 0$
$\Rightarrow k^2 > 20$
$\Rightarrow\text{k}>\sqrt{20}$ or $\text{k}<-\sqrt{20}$
$\Rightarrow\text{k}>2\sqrt5$ or $\text{k}<-2\sqrt5$

For equation $a x^2+b x+c=0$, it is given that $D=\left(b^2-4 a c\right) > 0$.
This means that the roots of the equation are real and unequal.

Since roots of the equation $ax^2 + bx + c = 0$ are equal,
$D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow b^2 = 4ac$
$\Rightarrow\text{c}=\frac{\text{b}^2}{\text{4a}}$

Perimeter of a rectangle $= 82m$
Let the breadth of the rectangle be $x m.$
Then, length of the rectangle $=\frac{\text{Perimeter}}{2}-\text{Breadth}$
$=\frac{82}{2}-\text{x}=(41-\text{x})\text{m}$
Now Area $= 400m^2$
$\Rightarrow$ Length $\times$ Breadth $= 400$
$\Rightarrow x(41 - x) = 400$
$\Rightarrow 41x - x^2 = 400$
$\Rightarrow x^2 - 41x + 400 = 0$
$\Rightarrow x^2 - 25x - 16x + 400 = 0$
$\Rightarrow x(x - 25) - 16(x - 25) = 0$
$\Rightarrow (x - 25)(x - 16) = 0$
$\Rightarrow x - 25 = 0$ or $x - 16 = 0$
$\Rightarrow x = 25$ or $x = 16$
Hence, the length is $25m$ and the breadth is $16m.$

Since the roots of the equation $x^2+ 6(k + 2)x + 9k = 0$ are equal,
$ D=0 $
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow[2(k+2)]^2-4 \times 1 \times 9 k=0 $
$ \Rightarrow 4\left(k^2+4 k+4\right)-36 k=0 $
$ \Rightarrow 4 k^2+16 k+16-36 k=0 $
$ \Rightarrow 4 k^2-20 k+16=0 $
$ \Rightarrow k^2-5 k+4=0 $
$ \Rightarrow k^2-4 k-k+4=0 $
$ \Rightarrow k(k-4)-1(k-4)=0 $
$ \Rightarrow(k-4)(k-1)=0 $
$ \Rightarrow k-4=0 \text { or } k-1=0 $
$ \Rightarrow k=4 \text { or } k=1$

Sum of the roots $= 6$
Product of the roots $= 6 × 6 = 36$
Required equation $= x^2- ($Sum of roots$)x +$ Product of roots $= 0$
$⇒x^2-6 x+6=0$

Since x = 3 is a solution of the equation $2x^2 + ax + 6 = 0$, we have
$2(2)^2 + a(2) + 6 = 0$
$\Rightarrow 8 + 2a + 6 = 0$
$\Rightarrow 2a = -14$
$\Rightarrow a = -7$

Sum of the roots $= 5 + (-2) = 3$
Product of the roots $= 5 \times (-2) = -10$
Required equation $= x^2 - ($Sum of roots$)x +$ Product of roots $= 0$
$\Rightarrow x^2- 3x - 10 = 0$

1. $\text{x}^2-3\sqrt{\text{x}}+2=0$ is not a quadratic equation, since it contains a term involving $\sqrt{\text{x}},$ i.e., $\text{x}^{\frac{1}{2}},$ where $\frac{1}{2}$ is not a integer.
2. $\text{x}+\frac{1}{\text{x}}=\text{x}^2$

3. $\text{x}^2+\frac{1}{\text{x}^2}=\text{5}$

4. $\text{2x}^2-\text{5x}=(\text{x}-1)^2$

$\Rightarrow 2x^2 - x - 6 = 0$
$\Rightarrow 2x^2 - 4x + 3x - 6 = 0$
$\Rightarrow 2x(x - 2) + 3(x - 2) = 0$
$\Rightarrow (x - 2)(2x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2x + 3 = 0$
$\Rightarrow x = 2$ or $\text{x}=\frac{-3}{2}$
Thus, the roots of the given equation are $2$ and $\frac{-3}{2}.$