Question #9b2b7

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Question #9b2b7

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Answer 1 · 2017-10-03T16:40:12

$20 x^{2} - 3 x - 9 = 0$ $20x^2-3x-9=0$

Explanation:

Let roots be $a = - \frac{3}{5}$ $a=-3/5$ and $b = \frac{3}{4}$ $b=3/4$
Sum of Roots $a + b = - \frac{3}{5} + \frac{3}{4} = 3 (\frac{1}{4} - \frac{1}{5}) = \frac{3}{20}$ $a+b=-3/5+3/4=3(1/4-1/5)=3/20$
Multiplication of roots $a b = - \frac{3}{5} \times \frac{3}{4} = - \frac{9}{20}$ $ab=-3/5xx3/4=-9/20$

If the roots of Quadratic equation is $a$ $a$ and $b$ $b$ then the equation will be

$x^{2} - (sum of roots) x + (Multiplication of roots) = 0$ $x^2-("sum of roots")x+("Multiplication of roots")=0$
$x^{2} - (a + b) x + a b = 0$ $x^2-(a+b)x+ab=0$
$x^{2} - \frac{3}{20} x - \frac{9}{20} = 0$ $x^2-3/20x-9/20=0$

Multiply whole equation by 20
$20 x^{2} - 3 x - 9 = 0$ $20x^2-3x-9=0$

Answer 2 · 2017-10-03T16:42:50

$20 x^{2} - 3 x - 9 = 0$ $20x^2-3x-9=0$

Explanation:

If we work backwards starting by letting x equal the two roots. Then make the equation equal to 0.

$x = - \frac{3}{5}$ $x=-3/5$
$x + \frac{3}{5} = 0$ $x+3/5=0$
$5 x + 3 = 0$ $5x+3=0$

$x = \frac{3}{4}$ $x=3/4$
$x - \frac{3}{4} = 0$ $x-3/4=0$
$4 x - 3 = 0$ $4x-3=0$

Now these equations equal 0 we can put them into the factorised form of a quadratic.

$(5 x + 3) (4 x - 3) = 0$ $(5x+3)(4x-3)=0$

Expand.

$20 x^{2} - 15 x + 12 x - 9 = 0$ $20x^2-15x+12x-9=0$
$20 x^{2} - 3 x - 9 = 0$ $20x^2-3x-9=0$

So $20 x^{2} - 3 x - 9 = 0$ $20x^2-3x-9=0$ is the quadratic with roots -3/5 and 3/4

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Question #9b2b7

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