Question

answer pls for 50 pts.
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Answer 1

Question:-

Find the nature of the roots of the quadratic equation, if the real roots exist. Find them,

a) 2x² - 3x + 5 = 0 (b) 3x² - 4√3x + 4 

Answer 1.

a) 2x² - 3x + 5 = 0

The above equation is in the form ax² + bx + c = 0

Here; a = 2, b = - 3 and c = 5

We know that,

D = b² - 4ac

$\implies\:\sf{(-3)^2\:-\:4(2)(5)}$

$\implies\:\sf{9\:-\:40}$

$\implies\:\sf{-\:31}$ (D < 0)

So, no real roots exist.

b) 3x² - 4√3x + 4

Here; a = 3, b = -4√3 and c = 4

D = b² - 4ac

$\implies\:\sf{(-4\sqrt{3})^2\:-\:4(3)(4)}$

$\implies\:\sf{48\:-\:48}$

$\implies\:\sf{0}$

Hence, it have two roots which are equal.

$\implies\:\sf{x\:=\:\frac{-b\pm \sqrt{D}}{2a}}$

$\implies\:\sf{x\:=\:\frac{-(-4\sqrt{3})\pm \sqrt{0}}{2(3)}}$

$\implies\:\sf{x\:=\:\frac{4\sqrt{3}\pm \sqrt{0}}{6}}$

$\implies\:\sf{x\:=\:\frac{2\sqrt{3}+0}{3}, \:\frac{2\sqrt{3}-0}{3}}$

The roots are 2√3/3 and 2√3/3.

$\rule{200}2$

Question:-

find the value of k so that they have equal roots

(a) 2x² + kx + 3 = 0 (b) kx(x - 2) + 6 = 0

Answer 2.

a) 2x² + kx + 3 = 0

The above equation is in the form ax² + bx + c = 0

Here, a = 2, b = k and c = 3

The above equation has two equal roots. So,

D = b² - 4ac = 0

$\implies\:\sf{(k)^2\:-\:4(2)(6)\:=\:0}$

$\implies\:\sf{k^2\:-\:24\:=\:0}$

$\implies\:\sf{k^2\:=\:24}$

$\implies\:\sf{k\:=\:\pm\sqrt{24}}$

$\implies\:\sf{k\:=\:\pm2\sqrt{6}}$

Value of k is ±2√6.

b) kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

Here; a = k, b = -2k and c = 6

The above equation has two equal roots. So,

D = b² - 4ac = 0

$\implies\:\sf{(-2k)^2\:-\:4(k)(6)\:=\:0}$

$\implies\:\sf{4k^2\:-\:24k\:=\:0}$

$\implies\:\sf{4k(k-6)\:=\:0}$

$\implies\:\sf{k\:=\:\frac{0}{4}, 6}$

$\implies\:\sf{k\:=\:0, \:6}$

Value of k is 6.

$\rule{200}2$

Question:-

Find the value of k for which the equation x² - 4x + k = 0 has distinct real roots.

Answer 3.

x² - 4x + k = 0

The above equation is in the form ax² - bx + c

Here; a = 1, b = -4 and c = k

Roots are distinct and real.

D > 0

b² - 4ac > 0

$\implies\:\sf{(-4)^2\:-\:4(1)(k)\:>\:0}$

$\implies\:\sf{16\:\:-\:4k\:>\:0}$

$\implies\:\sf{4k\:<\:16}$

$\implies\:\sf{k\:<\:4}$

Value of k < 4.

$\rule{200}2$

Question:-

Find the roots 1/x+4 -1/x-7 =11/30

Answer 4.

$\sf{\frac{1}{x+4}\:-\:\frac{1}{x-7}\:=\:\frac{11}{30}}$

$\implies\:\sf{\frac{(x-7)\:-\:(x+4)}{(x+4)(x-7)}\:=\:\frac{11}{30}}$

$\implies\:\sf{\frac{-11}{(x+4)(x-7)}\:=\:\frac{11}{30}}$

$\implies\:\sf{-11(30)\:=\:11[(x+4)(x-7)]}$

$\implies\:\sf{-330\:=\:11[(x^2\:-\:7x\:+\:4x\:-\:28)}$

$\implies\:\sf{-30\:=\:x^2\:-\:3x\:-\:28}$

$\implies\:\sf{x^2\:-\:3x\:+\:2=\:0}$

Using quadratic formula

Here; a = 1, b = -3 and c = 2

$\implies\:\sf{x\:=\:\frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}$

$\implies\:\sf{x\:=\:\frac{ - (-3)\pm \sqrt{ {(-3)}^{2} - 4(1)(2) } }{2(1)}}$

$\implies\:\sf{x\:=\:\frac{ 3\pm \sqrt{ 9 - 8} }{2}}$

$\implies\:\sf{x\:=\:\frac{3\:\pm\:1}{2}}$

$\implies\:\sf{x\:=\:\frac{3\:+\:1}{2}, \:\frac{3\:-\:1}{2}}$

$\implies\:\sf{x\:=\:\frac{4}{2}\:=\:2,\:\frac{2}{2}\:=\:1}$

Roots of the equations are 2, 1

Answer 2

Answer:

Question:-

Find the nature of the roots of the quadratic equation, if the real roots exist. Find them,

a) 2x² - 3x + 5 = 0 (b) 3x² - 4√3x + 4

Answer 1.

a) 2x² - 3x + 5 = 0

The above equation is in the form ax² + bx + c = 0

Here; a = 2, b = - 3 and c = 5

We know that,

D = b² - 4ac

\implies\:\sf{(-3)^2\:-\:4(2)(5)}⟹(−3)

2

−4(2)(5)

\implies\:\sf{9\:-\:40}⟹9−40

\implies\:\sf{-\:31}⟹−31 (D < 0)

So, no real roots exist.

b) 3x² - 4√3x + 4

Here; a = 3, b = -4√3 and c = 4

D = b² - 4ac

\implies\:\sf{(-4\sqrt{3})^2\:-\:4(3)(4)}⟹(−4

3

)

2

−4(3)(4)

\implies\:\sf{48\:-\:48}⟹48−48

\implies\:\sf{0}⟹0

Hence, it have two roots which are equal.

\implies\:\sf{x\:=\:\frac{-b\pm \sqrt{D}}{2a}}⟹x=

2a

−b±

D

\implies\:\sf{x\:=\:\frac{-(-4\sqrt{3})\pm \sqrt{0}}{2(3)}}⟹x=

2(3)

−(−4

3

)±

0

\implies\:\sf{x\:=\:\frac{4\sqrt{3}\pm \sqrt{0}}{6}}⟹x=

6

4

3

±

0

\implies\:\sf{x\:=\:\frac{2\sqrt{3}+0}{3}, \:\frac{2\sqrt{3}-0}{3}}⟹x=

3

2

3

+0

,

3

2

3

−0

The roots are 2√3/3 and 2√3/3.

Question:-

find the value of k so that they have equal roots

(a) 2x² + kx + 3 = 0 (b) kx(x - 2) + 6 = 0

Answer 2.

a) 2x² + kx + 3 = 0

The above equation is in the form ax² + bx + c = 0

Here, a = 2, b = k and c = 3

The above equation has two equal roots. So,

D = b² - 4ac = 0

\implies\:\sf{(k)^2\:-\:4(2)(6)\:=\:0}⟹(k)

2

−4(2)(6)=0

\implies\:\sf{k^2\:-\:24\:=\:0}⟹k

2

−24=0

\implies\:\sf{k^2\:=\:24}⟹k

2

=24

\implies\:\sf{k\:=\:\pm\sqrt{24}}⟹k=±

24

\implies\:\sf{k\:=\:\pm2\sqrt{6}}⟹k=±2

6

Value of k is ±2√6.

b) kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

Here; a = k, b = -2k and c = 6

The above equation has two equal roots. So,

D = b² - 4ac = 0

\implies\:\sf{(-2k)^2\:-\:4(k)(6)\:=\:0}⟹(−2k)

2

−4(k)(6)=0

\implies\:\sf{4k^2\:-\:24k\:=\:0}⟹4k

2

−24k=0

\implies\:\sf{4k(k-6)\:=\:0}⟹4k(k−6)=0

\implies\:\sf{k\:=\:\frac{0}{4}, 6}⟹k=

4

0

,6

\implies\:\sf{k\:=\:0, \:6}⟹k=0,6

Value of k is 6.

Question:-

Find the value of k for which the equation x² - 4x + k = 0 has distinct real roots.

Answer 3.

x² - 4x + k = 0

The above equation is in the form ax² - bx + c

Here; a = 1, b = -4 and c = k

Roots are distinct and real.

D > 0

b² - 4ac > 0

\implies\:\sf{(-4)^2\:-\:4(1)(k)\:>\:0}⟹(−4)

2

−4(1)(k)>0

\implies\:\sf{16\:\:-\:4k\:>\:0}⟹16−4k>0

\implies\:\sf{4k\:<\:16}⟹4k<16

\implies\:\sf{k\:<\:4}⟹k<4

Value of k < 4.

Question:-

Find the roots 1/x+4 -1/x-7 =11/30

Answer 4.

\sf{\frac{1}{x+4}\:-\:\frac{1}{x-7}\:=\:\frac{11}{30}}

x+4

1

−

x−7

1

=

30

11

\implies\:\sf{\frac{(x-7)\:-\:(x+4)}{(x+4)(x-7)}\:=\:\frac{11}{30}}⟹

(x+4)(x−7)

(x−7)−(x+4)

=

30

11

\implies\:\sf{\frac{-11}{(x+4)(x-7)}\:=\:\frac{11}{30}}⟹

(x+4)(x−7)

−11

=

30

11

\implies\:\sf{-11(30)\:=\:11[(x+4)(x-7)]}⟹−11(30)=11[(x+4)(x−7)]

\implies\:\sf{-330\:=\:11[(x^2\:-\:7x\:+\:4x\:-\:28)}⟹−330=11[(x

2

−7x+4x−28)

\implies\:\sf{-30\:=\:x^2\:-\:3x\:-\:28}⟹−30=x

2

−3x−28

\implies\:\sf{x^2\:-\:3x\:+\:2=\:0}⟹x

2

−3x+2=0

Using quadratic formula

Here; a = 1, b = -3 and c = 2

\implies\:\sf{x\:=\:\frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}⟹x=

2a

−b±

b

2

−4ac

\implies\:\sf{x\:=\:\frac{ - (-3)\pm \sqrt{ {(-3)}^{2} - 4(1)(2) } }{2(1)}}⟹x=

2(1)

−(−3)±

(−3)

2

−4(1)(2)

\implies\:\sf{x\:=\:\frac{ 3\pm \sqrt{ 9 - 8} }{2}}⟹x=

2

3±

9−8

\implies\:\sf{x\:=\:\frac{3\:\pm\:1}{2}}⟹x=

2

3±1

\implies\:\sf{x\:=\:\frac{3\:+\:1}{2}, \:\frac{3\:-\:1}{2}}⟹x=

2

3+1

,

2

3−1

\implies\:\sf{x\:=\:\frac{4}{2}\:=\:2,\:\frac{2}{2}\:=\:1}⟹x=

2

4

=2,

2

2

=1

Roots of the equations are 2, 1