MATHS M.C.Q (1 Marks)

2. 1

Solution:

Given equation: x² − bx + c = 0

Let $\alpha$ and $\alpha+1$ be the two consecutive roots of the equation.

Sum of the roots $=\alpha+\alpha+1=2\alpha+1$

Product of the roots $=\alpha(\alpha+1)=\alpha^2+\alpha$

So, sum of the roots $=2\alpha+1=\frac{-\text{Coeffecient of x}}{\text{Coeffecient of x}^2}=\frac{\text{b}}{1}=\text{b}$

Product of the roots $=\alpha^2+\alpha=\frac{\text{Constant term}}{\text{Coeffecient of x}^2}=\frac{\text{c}}{1}=\text{c}$

Now,$\text{b}^2-4\text{c}=(2\alpha+1)^2-4(\alpha^2+\alpha)=4\alpha^2+4\alpha+1-4\alpha^2-4\alpha=1$

1. 2

Solution:

(x² + 2x)² - (x + 1)² - 55 = 0

⇒ (x² + 2x + 1 - 1)² - (x + 1)² - 55 = 0

$\Rightarrow\Big\{(\text{x}+1)^2-1\Big\}^2-(\text{x}+1)^2-55=0$

$\Rightarrow\Big\{(\text{x}+1)^2\Big\}^2+1-3(\text{x}+1)^2-55=0$

$\Rightarrow\Big\{(\text{x}+1)^2\Big\}^2-3(\text{x}+1)^2-54=0$

Let p = (x + 1)²

⇒ p² - 3p - 54 = 0

⇒ p² - 9p + 6p - 54 = 0

⇒ (p + 6) (p - 9) = 0

⇒ p = 9 or p = -6

Rejecting p = −6

⇒ (x + 1)² = 9

⇒ x² + 2x - 8 = 0

⇒ (x + 4) (x - 2) = 0

⇒ x = 2, x = -4

2. $-\frac{3}{7}$

Solution:

Given equation: 4x² + 3x + 7 = 0

Also, $\alpha$ and $\beta$ are the roots of the equation.

Then, sum of the roots = $\alpha$ + $\beta$ $=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\frac{3}{4}$

Product of the roots $=\alpha\beta=\frac{\text{constant term}}{\text{Coefficient of x}^2}=\frac{7}{4}$

$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-\frac{3}{4}}{\frac{7}{4}}=-\frac{3}{7}$

1. $\frac{49}{4}$

Solution:

It is given that, 4 is the root of the equation x² + px + 12 = 0.

$\therefore$ 16 + 4p + 12 = 0

⇒ p = -7

It is also given that, the equation x² + px + q = 0 has equal roots. So, the discriminant of:

x² + px + q = 0 will be zero.

$\therefore$ p² - 4q = 0

⇒ 4q = (-7)² = 49

$\Rightarrow\text{q}=\frac{49}{4}$

2. $2\pm\sqrt{12}$

Solution:

Since the equation has real roots.

⇒ D = 0

⇒ b² - 4ac = 0

⇒ K² - 4 (1)(K + 2) = 0

⇒ K² - 4K - 8 = 0

$\Rightarrow\text{K}=\frac{4\pm\sqrt{16-4(1)(-8)}}{2(1)}$

$\Rightarrow\text{K}=\frac{4\pm2\sqrt{12}}{2}$

$\Rightarrow\text{K}=2\pm\sqrt{12}$

2. p² − 4q = 1

Solution:

Given equation: x² + px + q = 0

Also, $\alpha$ and $\beta$ are the roots of the equation such that $\alpha-\beta=1$

Sum of the roots $=\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\Big(\frac{-\text{p}}{1}\Big)=\text{p}$

Product of the roots $=\alpha\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}=\text{q}$

$\therefore(\alpha+\beta)^2-(\alpha-\beta)^2=4\alpha\beta$

⇒ p² - 1 = 4q

⇒ p² - 4q =1.

1. p = 1, q = −2

Solution:

It is given that, p and q $(\text{P}\neq0,\ \text{q}\neq0)$are the roots of the equation x2 + px + q = 0

$\therefore$ Sum of roots = p + q = −p

⇒ 2p + q = 0 ...(1)

Product of roots = pq = q

⇒ q (p − 1) = 0

⇒ p = 1, q = 0

 Now, substituting p = 1 in (1), we get,

2 + q = 0

⇒ q = −2

3. [-4, -3]

Solution:

The roots of the quadratic equation x² − (m + 1) x + m + 4 = 0 will be real, if its discriminant is greater than or equal to zero.

$\therefore$ (m + 1)² - 4 (m + 4) > 0

⇒ (m - 5) (m + 3) > 0

⇒ m < -3 or m > 5 ...(1)

It is also given that, the roots of x² − (m + 1) x + m + 4 = 0 are negative. So, the sum of the roots will be negative.

$\therefore$ Sum of the roots < 0

⇒ m + 1 < 0 ....(2)

and product of zeros > 0

⇒ m + 4 > 0 ...(3)

From (1), (2) and (3), we get,

[−4, −3]

2. -1

Solution:

Let a be the common roots of the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$

Therefore

$\alpha^2+2\text{a}+3\lambda=0\ ...(1)$

$2\alpha^2+3\alpha+5\lambda=0\ ...(2)$

Solving (1) and (2) by cross multiplication, we get

$\frac{\alpha^2}{10\lambda-9\lambda}=\frac{\alpha}{6\lambda-5\lambda}=\frac{1}{3-4}$

$\Rightarrow\text{a}^2=-\lambda,\alpha=-\lambda$

$\Rightarrow-\lambda=\lambda^2$

$\Rightarrow\lambda=-1$

3. $\frac{\text{b}}{\text{ac}}$

Solution:

Given equation: ax² + bx + c = 0

Also, $\alpha$ and $\beta$ are the roots of the given equation.

Then, sum of the roots $=\alpha+\beta=-\frac{\text{b}}{\text{a}}$

Product of the roots $=\alpha\beta=\frac{\text{c}}{\text{a}}$

$\therefore\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}=\frac{\text{a}\beta+\text{b}+\text{a}\alpha+\text{b}}{(\text{a}\alpha+\text{b})(\text{a}\beta+\text{b})}$

$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}\alpha+\text{ab}\beta+\text{b}^2}$

$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}(\alpha+\beta)+\text{b}^2}$

$=\frac{\text{a}\big(-\frac{\text{b}}{\text{a}}\big)+2\text{b}}{\text{a}^2\big(\frac{\text{c}}{\text{a}}\big)+\text{ab}\big(-\frac{\text{b}}{\text{a}}\big)+\text{b}^2}$

$=\frac{\text{b}}{\text{ac}}$

3. 5, -7

Solution:

The given equation is kx² + 1 = kx + 3x - 11x² which can be written as.

kx² + 11x² - kx - 3x + 1 = 

⇒ (k + 11)x² - (k + 3)x + 1 = 0

For equal and real roots, the discriminant of (k + 11)x² - (k + 3)x + 1 = 0.

$\therefore$ (k + 3)² -4(k + 11) = 0

⇒ k² + 2k - 35 = 0

⇒ (k - 5)(k + 7) = 0

⇒ k = 5, -7

Hence, the equation has real and equal roots when k = 5, -7.

1. q² − p²

Solution:

Given: $\alpha$ and $\beta$ are the roots of the equation x² + px + 1 = 0.

Also, $\gamma$ and $\delta\gamma$ and $\delta$ are the roots of the equation x² + qx + 1 = 0

Then, the sum and the product of the roots of the given equation are as follows:

$\alpha+\beta=-\frac{\text{p}}{1}=-\text{p}$

$\alpha\beta=\frac{1}{1}=1$

$\gamma+\delta=-\frac{\text{q}}{1}=-\text{P}$

$\gamma\delta=\frac{1}{1}=1$

Moreover, $(\gamma-\delta)^2=\gamma^2+\delta^2+2\gamma\delta$

$\Rightarrow\gamma^2+\delta^2=\text{q}^2-2$

$\therefore(\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta+\delta)=(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$

$=(\alpha\beta-\alpha\gamma-\beta\gamma+\gamma^2)(\alpha\beta+\alpha\gamma+\beta\delta+\delta^2)$

$=[\alpha\beta-\gamma(\alpha+\beta)+\gamma^2][\alpha\beta+\delta(\alpha+\beta)+\delta^2])$

$=(1-\gamma(-\text{p})+\gamma^2)(1+\delta(-\text{p}+\delta^2))$

$=(1+\gamma\text{p}+\gamma^2)(1-\delta\text{p}+\delta^2)$

$=(1+\gamma\text{p}+\gamma^2)(1-\delta\text{p}+\delta^2)$

$=1-\text{p}\delta+\delta^2+\text{p}\gamma-\text{p}^2\gamma\delta+\text{p}\gamma\delta^2+\gamma^2-\text{p}\delta\text{y}^2+\gamma^2\delta^2$

$=1-\text{p}\delta+\text{p}\gamma+\delta^2-\text{p}^2\gamma\delta+\text{p}\gamma\delta^2+\gamma^2=\text{p}\delta\gamma^2+\gamma^2\delta^2$

$=1-\text{p}(\delta-\gamma)-\text{p}^2\gamma\delta+\text{p}\gamma\delta(\delta-\gamma)+(\gamma^2+\delta)+1$

$=1-\text{p}^2+(\delta-\gamma)\text{p}(\delta-\gamma)+(\gamma^2+\delta^2)+1$

$=-\text{P}^22+(\delta-\gamma)\text{p}(1-1)+\text{q}^2$

$=\text{q}^2-\text{p}^2$

3. 24

Solution:

Let $\alpha$ be the common roots of the equations

x² - 11x + a = 0 and x² - 14x + 2a = 0

Therefore,

$\alpha^2-11\alpha+\alpha=0\ ...(1)$

$\alpha^2-14\alpha+2\alpha=0\ ...(2)$

Solving (1) and (2) by cross multiplication, we get,

$\frac{\alpha^2}{-22\alpha+14\alpha}=\frac{\alpha}{\alpha-2\alpha}=\frac{1}{-14+11}$

$\Rightarrow\alpha^2=\frac{-22\alpha+14\alpha}{-14+11},\alpha=\frac{\alpha-2\alpha}{-14+11}$

$\Rightarrow\alpha^2=\frac{-8\alpha}{-3}=\frac{8\alpha}{3},\alpha=\frac{-\alpha}{-3}=\frac{\alpha}{3}$

$\Rightarrow\Big(\frac{\alpha}{3}\Big)^2=\frac{8\alpha}{3}$

$\Rightarrow\alpha^2=24\alpha$

$\Rightarrow\alpha^2-24\alpha=0$

$\Rightarrow\alpha(\alpha-24)=0$

$\Rightarrow\alpha=0$ or $\alpha=24$

2. 0

Solution:

Let P = |x|

⇒ p² + p - 6 = 0

⇒ p² + 3p - 2p - 6 = 0

⇒ (p + 3) (p - 2) = 0

⇒ p = -3, 2

Also, |x| = p

⇒ |x| = 2, or |x| = -3

Modules can not be negative,

$\therefore|\text{x}| = 2$ 

$\Rightarrow\text{x} = \pm 2$

⇒ x = 2 or −2

Sum of the roots of x is 0.

1. $\text{K}\in\Big[\frac{1}{3,3}\Big]$

Solution:

$\text{K}=\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}$

⇒ kx² + kx + K = x² - x + 1

⇒ (k - 1)x² + (k + 1) x + k - 1 = 0

For real values of x, the discriminant of (k - 1)x² + (k + 1) x + k - 1 = 0 should be greater than or equal to zero.

$\therefore$ If $\text{k}\neq1$

(k + 1)² - 4(k - 1) (k - 1) > 0

$\Rightarrow(\text{k}+1)^2-\big\{2(\text{k}-1)\big\}^2>0$

⇒ (3k - 1) (k - 3) < 0

$\Rightarrow\frac{1}{3}<\text{K}<3$

And if k = 1, then,

x = 0, which is real ...(ii)

So, from (i) and (ii), we get,

$\text{k}\in\Big[\frac{1}{3},3\Big]$

4. $\frac{-3}{7}$

Solution:

Given equation: 4x² + 3x + 7 = 0

Also, $\alpha$ and $\beta$ are the roots of the equation.

Sum of the roots $=\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\frac{3}{4}$

Product of the roots $=\alpha\beta=\frac{\text{Coefficient term}}{\text{Coefficient of x}^2}=\frac{7}{4}$

$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-\frac{3}{4}}{\frac{7}{4}}=-\frac{3}{7}$

2. 1

Solution:

$\frac{(\text{x}+2)(\text{x}-5)}{(\text{x}+3)(\text{x}+6)}=\frac{\text{x}-2}{\text{x}+4}$

⇒ (x² - 3x - 10) (x + 4) = (x² + 3x - 18) (x - 2)

⇒ x³ + 4x² - 3x² - 12x - 10x - 40 = x³ - 2x² + 3x² - 6x - 18x + 36

⇒ x² - 22x - 40 = x² - 24x + 36

⇒ 2x = 76

⇒ x = 38

Hence, the equation has only 1 root.

3. qx² − px + 1 = 0

Solution:

Given equation: x² + px + q = 0

Also, $\alpha$ and $\beta$ are the roots of the given equation.

Then, sum of the roots $=\alpha+\beta=-\text{p}$

Product of the roots $=\alpha\beta=\text{q}$

Now, for roots $-\frac{1}{\alpha,}-\frac{1}{\beta},$ we have:

Sum of the roots = $-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha\beta}=-\Big(\frac{-\text{p}}{\text{q}}\Big)=\frac{\text{p}}{\text{q}}$

Product of the roots = $\frac{1}{\alpha\beta}=\frac{1}{\text{q}}$

Hence, the equation involving the roots $-\frac{1}{\alpha},-\frac{1}{\beta}$ is as follows:

$\text{x}^2+(\alpha+\beta)\text{x}+\alpha\beta=0$

$\Rightarrow\text{x}^2-\frac{\text{p}}{\text{q}}\text{x}+\frac{1}{\text{q}}=0$

$\Rightarrow\text{qx}^2-\text{px}+1=0$

3. 1 - c

Solution:

Given equation:

x² − p(x + 1) − c = 0

or x² − px − p − c = 0

Also $\alpha $ and $\beta$ are the roots of the equation.

Sum of the roots $=\alpha+\beta=\text{p}$

Product of the roots $=\alpha\beta=-(\text{c}+\text{p})$

Then, $(\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1$

$=-(\text{c}+\text{p})+\text{p}+1$

$=1-\text{c}$

$=-\text{c}-\text{p}+\text{p}+1$

3. -1

Solution:

Given equation: x² + x + 1 = 0

Also, a and b are the roots of the given equation.

Sum of the roots $=\text{a}+\text{b}=\frac{-\text{Co-efficient of x}}{\text{C-oefficient of x}^2}=-\frac{1}{1}=-1$

Product of the roots $=\text{ab}=\frac{\text{constant term}}{\text{Coefficient of x}}=\frac{1}{1}=1$

$\therefore$ (a + b)² = a² + b² + 2ab

⇒ (-1)² = a² + b² + 2 × 1

⇒ 1 - 2 = a² + b²

⇒ a² + b² = -1

4. 7

Solution:

The roots of the quadratic equation x² + 5x + k=0 will be imaginary if its discriminant is less than zero.

$\therefore25-4\text{0k}<0$

$\Rightarrow\text{k}>\frac{25}{4}$

Thus, the minimum integral value of k for which the roots are imaginary is 7.