Question

I. Solve the following quadratic equations by factorization:
i) 2x^2 + 5x +3 =0
ii) x ^2 -3x +2 =0
iii) 2x^ 2 – 7x =0
iv) x ^2 +5x -6 =0
v) 2x^ 2 +6x+8 =0
vi) 3x ^2 +2x -1 =0
vii) 2/5x^ 2 – x − 3/5 =0
viii) 6x ^2 +x -12 =0
ix) 5x ^2 +4x=0
x) √7x^ 2 – 6x -13 √7 =0

Answer 1

Solutions

i) 2x² + 5x + 3 = 0

By Factorizing we have

$2 {x}^{2} + 5x + 3 = 0 \\ \implies2 {x}^{2} + 2x + 3x + 3 = 0 \\ \implies2x(x + 1) + 3(x + 1) = \\ \implies(x + 1)(2x + 3)$

Now the solutions are :

$x + 1 = 0 \: \: \: and \: \: \: 2x + 3 = 0 \\ \implies \bold{x = - 1} \: \: and \implies \bold{x = \frac{- 3}{2} }$

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ii) x² - 3x + 2 = 0

By factorizing we have

${x}^{2} - 3x + 2 = 0 \\ \implies {x}^{2} - x - 2x + 2 = 0 \\ \implies x(x - 1) - 2(x - 1) = 0 \\ \implies(x - 1)(x - 2) = 0$

Now the solutions are :

$x - 1 = 0 \: \: \: and \: \: x - 2 = 0 \\ \implies \bold{x = 1} \: and \: \implies \bold{x = 2}$

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iii)2x² - 7x = 0

By Factorizing we have

$2 {x}^{2} - 7x = 0 \\ \implies x(2x - 7) = 0$

Now the solutions are :

$x = 0 \: and \: 2x - 7 = 0 \\ \implies \bold{x = 0} \: and \implies \bold{x = \frac{7}{2} }$

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iv) x² + 5x - 6 = 0

By Factorizing we have

${x}^{2} - 5x + 6 = 0 \\ \implies {x}^{2} - 2x - 3x + 6 = 0 \\ \implies x( x - 2) - 3(x - 2) = 0\\ \implies(x - 2)(x - 3) = 0$

Now the solutions are :

$x - 2 = 0 \: \: and \: \: x - 3 = 0 \\ \implies \bold{x = 2} \: and \implies \bold{x = 3 }$

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v) 2x² + 6x + 8 = 0

By Factorizing we have

$2 {x}^{2} + 6x + 8 = 0 \\ \implies {x}^{2} + 3x + 4 = 0$

By applying quadratic formula

x =[-b ±√(b² -4ac)]/2a

→ x =[ -3±√(9 - 4×1×4) ]/2×1

→ x = (-3 ± √-7)/2

So the solutions are

$\bold{x = \frac{ - 3 + \sqrt{7}i }{2} } \: and \: \: \: \bold{ \frac{ - 3 + \sqrt{7}i }{2}}$

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vi) 3x² +2x - 1 = 0

By Factorizing we have

$3 {x}^{2} + 2x - 1 = 0 \\ \implies3 {x}^{2} - 3x + x - 1 = 0 \\ \implies3x(x - 1) + 1(x - 1) = 0 \\ \implies(x - 1)(3x + 1) = 0$

Now the solutions are:

$x - 1 = 0 \: and \: \: \: 3x + 1 = 0 \\ \implies \bold{x = 1 } \: and \implies \bold{x = \frac{ - 1}{3} }$

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vii) 2/5x² - x - 3/5 = 0

$\frac{2}{5} {x}^{2} - x - \frac{3}{5} = 0 \\ \implies2 {x}^{2} - 5x - 3 = 0 \\ \implies2 {x}^{2} + x - 6x - 3 = 0 \\ \implies x(2x + 1) - 3(2x + 1) = 0 \\ \implies(2x + 1)(x - 3) = 0$

Now the solutions are :

$2x + 1 = 0 \: \: and \: \: x - 3 = 0 \\ \implies \bold{x = - \frac{1}{2} } and\implies \bold{x = 3}$

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viii) 6x² +x -12 = 0

By Factorizing we have

$6 {x}^{2} + x - 12 = 0 \\ \implies6 {x}^{2} + 9x - 8x - 12 = 0 \\ \implies3x(2x + 3) - 4(2x + 3) = 0 \\ \implies(2x + 3)(3x - 4) = 0$

Now the solutions are :

$2x + 3 = 0 \: \: and \: 3x - 4 = 0 \\ \implies \bold{x = \frac{ - 3}{2} } \: and \implies \bold{x = \frac{4}{3} }$

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ix) 5x² + 4x = 0

By Factorizing we have

$5 {x}^{2} + 4x = 0 \\ \implies x(5x + 4) = 0$

Now the solutions are:

$x = 0 \: \: and \: \: 5x + 4 = 0 \\ \implies \bold{x = 0} \: \: and \implies \bold{x = \frac{ - 4}{5} }$

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x)√7x² - 6x - 13√7 = 0

By Factorizing we have

$\sqrt{7} {x}^{2} - 6x - 13 \sqrt{7} = 0 \\ \implies \sqrt{7} {x}^{2} + 7x - 13x - 13 \sqrt{7} = 0 \\ \implies \sqrt{7} x(x + \sqrt{7} ) - 13(x + \sqrt{7} ) = 0 \\ \implies(x + \sqrt{7} )( \sqrt{7} x - 13) = 0$

Now the solutions are :

$x + \sqrt{7} = 0 \: \: and \: \: \: \sqrt{7} x - 13 = 0 \\ \implies \bold{x = - \sqrt{7} } \implies \bold{x = \frac{13}{ \sqrt{7} } }$

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