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marissa [1.9K]
3 years ago
5

Write the quadratic equation whose roots are -3 and 2 , and whose leading coefficient is 5

Mathematics
1 answer:
KiRa [710]3 years ago
5 0

Step-by-step explanation:

Since, roots are - 3 & 2

Therefore, (x + 3) & (x - 2) would be factors.

Leading coefficient is 5

Hence, required quadratic equation is:

5(x + 3)(x - 2) = 0 \\   \therefore \:  5( {x}^{2}  - 2x + 3x - 6)  = 0\\  \therefore \:  5( {x}^{2}  + x - 6)  = 0\\ \therefore \: 5 {x}^{2}  + 5x - 30  = 0\:  \\ is \: the \: required \:   quadratic \: equation.

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What are the two missing sides of the triangle ??
igor_vitrenko [27]
Remark
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.

Givens
a = x
b = x + 4
c = 20

Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2

Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20  Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0  
2x^2 + 8x - 4 = 0         Divide through by 2
x^2 + 4x - 2 = 0

Use the quadratic formula
a = 1
b = 4
c = - 2

\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} 
 
\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} + 4(1)(2) } }{2} 


From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6)  This is an extraneous root. No line can be minus.

x1 = + 0.4495
x2 = x + 4 = 4.4495
6 0
3 years ago
Factor this polynomial completely.<br> 12x2+x-6
jonny [76]

Answer:

(

4

x

+

3

)

(

3

x

−

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Step-by-step explanation:

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Answer:

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open the brackets:

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bring on one side of the equal sign all terms containg m, to make it the subject:

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Answer:

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