Question

Find the roots / zeros of the following quadratic equations using any method 1. x( x-5)= 0 2. x^2+8x-20= 0 3.x^2-10x=56 4. 2x^2+28x+48=0 5. x^2+2x-3=0 6.(x-5)(x+3)=0 7. x^2-6x+5=0 8. 3x^2-12x-15=0 9. (2x+3)(x-4)=0 10.x^2-12x-85=0
answer and with solution po

Answer 1

Answer:

ax2 + bx + c = 0 where a = 6 , b = 3 and c = 5. ... Find k if x = 3 is a root of equation kx2 – 10x + 3 = 0 ... (5) 2x2 – 2x + 12 = 0 ... Now b2-4ac=(-4)2-4×5×(-2)=16+40=56

Answer 2

Solution 1 :-

=> x ( x - 5 ) = 0 .

=> x = 0

or

=> x - 5 = 0

=> x = 5

$\large\underline{\boxed{\green{\bf x = 0,5 }}}$

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Solution 2 :-

=> x² + 8x - 20 = 0

=> x² + 10x - 2x - 20 = 0

=> x ( x + 10 ) - 2 ( x + 10 ) = 0

=> ( x + 10 )( x - 2 ) = 0

=> x = 2 , (-10)

$\large\underline{\boxed{\green{\bf x = 2 , (-10) }}}$

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Solution 3 :-

=> x² - 10x = 56 .

=> x² - 10x - 56 = 0 .

=> x² - 14x + 4x - 56 = 0.

=> x ( x - 14 ) + 4 ( x - 14 ) = 0.

=> ( x - 14 ) ( x + 4 ) = 0.

=> x = 14 , (-4)

$\large\underline{\boxed{\green{\bf x = 14,(-4) }}}$

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Solution 4 :-

=> 2x² + 28x + 48 = 0.

=> 2 ( x² + 14x + 24 ) = 0

=> x² + 14x + 24 = 0

=> x² + 12x + 2x + 24 = 0

=> x ( x + 12 ) + 2 ( x + 12 ) = 0

=> ( x + 12 ) ( x + 2 ) = 0.

=> x = (-12) , (-2)

$\large\underline{\boxed{\green{\bf x = (-12),(-2) }}}$

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Solution 5 :-

=> x² + 2x - 3 = 0 .

=> x² + 3x - x - 3 = 0.

=> x ( x + 3 ) -1( x + 3 ) = 0.

=> ( x + 3 ) ( x - 1 ) =0

=> x = (-3) , 1

$\large\underline{\boxed{\green{\bf x = (-3),1 }}}$

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