Question

1.Find the roots of the following quadratic equations by factorisation:
i )x²-3x - 10=0
ii)2x² +x-6=0
iii)root2x²+7x+5root2=0
iv)2x²-x+ 1/8 = 0
v)100x² - 20x + 1 = 0
vi)x(x- 4 ) = 12
Vii )3x² - 5x + 2 = 0
viii) x-3/2= 2 (x is not equal to 0)
ix )3(x-4)² – 5(x -4)=12​

Answer 1

Answer:

x2 – 3x – 10

= x2 - 5x + 2x - 10

= x(x - 5) + 2(x - 5)

= (x - 5)(x + 2)

Roots of this equation are the values for which (x - 5)(x + 2) = 0

∴ x - 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

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(ii) 2x2 + x – 6

= 2x2 + 4x - 3x - 6

= 2x(x + 2) - 3(x + 2)

= (x + 2)(2x - 3)

Roots of this equation are the values for which (x + 2)(2x - 3) = 0

∴ x + 2 = 0 or 2x - 3 = 0

⇒ x = -2 or x = 3/2

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(iii) √2 x2 + 7x + 5√2

= √2 x2 + 5x + 2x + 5√2

= x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

Roots of this equation are the values for which (√2x + 5)(x + √2) = 0

∴ √2x + 5 = 0 or x + √2 = 0

⇒ x = -5/√2 or x = -√2

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(iv) 2x2 – x + 1/8

= 1/8 (16x2 - 8x + 1)

= 1/8 (16x2 - 4x -4x + 1)

= 1/8 (4x(4x - 1) -1(4x - 1))

= 1/8(4x - 1)2

Roots of this equation are the values for which (4x - 1)2 = 0

∴ (4x - 1) = 0 or (4x - 1) = 0

⇒ x = 1/4 or x = 1/4

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(v) 100x2 – 20x + 1

= 100x2 – 10x - 10x + 1

= 10x(10x - 1) -1(10x - 1)

= (10x - 1)2

Roots of this equation are the values for which (10x - 1)2 = 0

∴ (10x - 1) = 0 or (10x - 1) = 0

⇒ x = 1/10 or x = 1/10

Step-by-step explanation:

I hope this will be helpful to you.

I hope this will be helpful to you. Now, I guess you can do remaining parts

Answer 2

Required Solution:-

i) $\sf{x^2 - 3x - 10 = 0}$

We know,

When an equation is in the form $\sf{ax^2 + bx + c = 0}$, the formula used to find the roots of the equation is,

$\sf{\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Here, the given equation is also in the $\sf{ax^2 + bx + c = 0}$, where:-

a = 1

b = -3

c = -10

Now,

Putting the values in the formula,

$\sf{\dfrac{-(-3) \pm \sqrt{(-3)^2 - 4\times (1)\times (-10)}}{2\times1}}$

= $\sf{\dfrac{3 \pm \sqrt{9 + 40}}{2}}$

= $\sf{\dfrac{3\pm \sqrt{49}}{2}}$

= $\sf{\dfrac{3\pm 7}{2}}$

= $\sf{\dfrac{3+7}{2}\:\:and\:\:\dfrac{3-7}{2}}$

= $\sf{\dfrac{10}{2}\:\:and\:\:\dfrac{-4}{2}}$

= $\sf{5\:\:and\:\:-2}$

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ii) $\sf{2x^2 + x - 6 = 0}$

Here,

a = 2

b = 1

c = -6

Now,

Putting the values in the formula,

= $\sf{\dfrac{-(1) \pm \sqrt{(1)^2 - 4\times2\times (-6)}}{2\times2}}$

= $\sf{\dfrac{-1 \pm \sqrt{1 + 48}}{4}}$

= $\sf{\dfrac{-1\pm \sqrt{49}}{4}}$

= $\sf{\dfrac{-1 \pm 7}{4}}$

= $\sf{\dfrac{-1+7}{4}\:\:and\:\:\dfrac{-1-7}{4}}$

= $\sf{\dfrac{6}{4}\:\:and\:\:\dfrac{-8}{4}}$

= $\sf{\dfrac{3}{2}\:\:and\:\:-2}$

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iii) $\sf{\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0}$

Here,

a = √2

b = 7

c = 5√2

Putting the values in the formula,

= $\sf{\dfrac{-(7) \pm \sqrt{(7)^2 - 4\times \sqrt{2}\times 5\sqrt{2}}}{2\times\sqrt{2}}}$

= $\sf{\dfrac{-7 \pm \sqrt{49 - 40}}{2\sqrt{2}}}$

= $\sf{\dfrac{-7 \pm \sqrt{9}}{2\sqrt{2}}}$

= $\sf{\dfrac{-7\pm 3}{2\sqrt{2}}}$

= $\sf{\dfrac{-7+3}{2\sqrt{2}}\:\:and\:\:\dfrac{-7-3}{2\sqrt{2}}}$

= $\sf{\dfrac{-4\times\sqrt{2}}{2\sqrt{2}\times\sqrt{2}}\:\:and\:\:\dfrac{-10\times\sqrt{2}}{2\sqrt{2}\times\sqrt{2}}}$

= $\sf{\dfrac{2\sqrt{2}}{2}\:\:and\:\:\dfrac{-10\sqrt{2}}{2\times2}}$

= $\sf{\sqrt{2}\:\:and\:\:\dfrac{-5\sqrt{2}}{2}}$

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iv) $\sf{2x^2 -x + \dfrac{1}{8} = 0}$

Here,

a = 2

b = -1

c = $\sf{\dfrac{1}{8}}$

Putting the values in the formula,

$\sf{\dfrac{-(-1) \pm \sqrt{(-1)^2 - 4\times2\times\dfrac{1}{8}}}{2\times2}}$

= $\sf{\dfrac{1\pm \sqrt{1 - 8\times\dfrac{1}{8}}}{4}}$

= $\sf{\dfrac{1\pm \sqrt{1-1}}{4}}$

= $\sf{\dfrac{1+0}{4}\:\:and\:\:\dfrac{1-0}{4}}$

= $\sf{\dfrac{1}{4}\:\:and\:\:\dfrac{1}{4}}$

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v) $\sf{100x^2 - 20x + 1 = 0}$

Here,

a = 100

b = -20

c = 1

Putting the values in the formula,

$\sf{\dfrac{-(-20) \pm \sqrt{(-20)^2 - 4\times100\times1}}{2\times100}}$

= $\sf{\dfrac{20 \pm \sqrt{400 - 400}}{200}}$

= $\sf{\dfrac{20 \pm 0}{200}}$

= $\sf{\dfrac{20}{200}\:\:and\:\:\dfrac{20}{200}}$

= $\sf{\dfrac{1}{10}\:\:and\:\:\dfrac{1}{10}}$

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vi) $\sf{x(x-4) = 12}$

In this question we need to first remove the bracket.

$\sf{x^2 - 4x - 12 = 0}$

Now,

a = 1

b = -4

c = -12

Putting the values in the formula,

= $\sf{\dfrac{-(-4)\pm \sqrt{(-4)^2 - 4\times1\times(-12)}}{2\times1}}$

= $\sf{\dfrac{4\pm \sqrt{16+48}}{2}}$

= $\sf{\dfrac{4\pm \sqrt{64}}{2}}$

= $\sf{\dfrac{4\pm 8}{2}}$

= $\sf{\dfrac{4-8}{2}\:\:and\:\:\dfrac{4+8}{2}}$

= $\sf{\dfrac{-4}{2}\:\:and\:\:dfrac{12}{2}}$

= $\sf{-2\:\:and\:\:6}$

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vii) $\sf{3x^2 - 5x + 2 = 0}$

Here,

a = 3

b = -5

c = 2

Putting the values in the formula,

$\sf{\dfrac{-(-5) \pm \sqrt{(-5)^2 - 4\times3\times2}}{2\times3}}$

= $\sf{\dfrac{5\pm \sqrt{25 - 24}}{6}}$

= $\sf{\dfrac{5\pm \sqrt{1}}{6}}$

= $\sf{\dfrac{5+1}{6}\:\:and\:\:\dfrac{5-1}{6}}$

= $\sf{\dfrac{6}{6}\:\:and\:\:\dfrac{4}{6}}$

= $\sf{1\:\:and\:\:\dfrac{2}{3}}$

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viii) $\sf{x-\dfrac{3}{2} = 2}$

Here as $\sf{x\neq 0}$ and the equation is not in the form of $\sf{ax^2 + bx + c}$

Therefore,

$\sf{x - \dfrac{3}{2} = 2}$

= $\sf{x = 2+\dfrac{3}{2}}$

= $\sf{x = \dfrac{4+3}{2}}$

= $\sf{x = \dfrac{7}{2}}$

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ix) $\sf{3(x-4)^2 - 5(x-4) = 12}$

$\sf{(x-4)[3(x-4) - 5] = 12}$

= $\sf{(x-4)[3x - 12 - 5] = 12}$

= $\sf{(x-4)[3x-17] = 12}$

= $\sf{3x^2 - 17x - 12x + 58 = 12}$

= $\sf{3x^2 -29x + 68 - 12 = 0}$

= $\sf{3x^2 - 29x + 56 = 0}$

Now,

Here,

a = 3

b = -29

c = 56

Putting the value in the formula,

= $\sf{\dfrac{-(-29)\pm \sqrt{(-29)^2 - 4\times3\times56}}{2\times3}}$

= $\sf{\dfrac{29\pm \sqrt{841 - 672}}{6}}$

= $\sf{\dfrac{29\pm \sqrt{169}}{6}}$

= $\sf{\dfrac{29\pm 13}{6}}$

= $\sf{\dfrac{29+13}{6}\:\:and\:\:\dfrac{29-13}{6}}$

= $\sf{\dfrac{42}{6}\:\:and\:\:\dfrac{16}{6}}$

= $\sf{7\:\:and\:\:\dfrac{8}{3}}$

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