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3 years ago

The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.

Mathematics

2 answers:

RideAnS [48]3 years ago

8 0

Answer:

Quadratic equation is x² + x - 12 = 0.

Step-by-step explanation:

Given : The solutions to a certain quadratic equation are x = -4 and x = 3.

To find : Write the equation in standard form below.

Solution : We have given x = -4 and x = 3.

We can write the solution as x +4 = 0

x - 3 = 0.

Both the solution have GCF is 1

Then solution are (x +4)(x -3) = 0.

On distributing x over (x-3) and +4 over (x-3).

x(x-3) +4(x-3) = 0.

x² -3x +4x -12 = 0.

x² + x - 12 = 0.

Then quadratic equation is x² + x - 12 = 0.

Therefore, Quadratic equation is x² + x - 12 = 0.

saw5 [17]3 years ago

7 0

Answer:

$x^2 + x -12$

Step-by-step explanation:

If the solutions are x=-4 and x=3, then it must have factors (x+4)(x-3). Assuming there is no GCF or leading coefficient other than 1, multiply with FOIL to find the standard form.

$(x+4)(x-3)\\x^2 + 4x -3x -12\\x^2 + x -12$

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Read 2 more answers

A coin will be tossed three times, and each toss will be recorded as heads (H) or tails (T). Give the sample space describing al

cricket20 [7]

Answer:

Sample space:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Third toss is heads:

{HHT, HTT, THT, TTT}

First toss is heads, second tails and third heads:

{HTH}

Explanation:

In order to find the sample space we need to write all the possible outcomes from tossing the coin three times.You can start by writting all the possible outcomes for the first toss being a head and finish with all the possible outcomes for the first toss being Tails. The number of elements in the sample space can be found by multiplying the possible outcomes of each toss together, so 2*2*2=8

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

For the third toss being heads we will have one restriction this time. Which is that the third toss is Tails. You can take the elements from the sample space where the third element is a tails. In this case, the number of elements can be calculated by multiplying the possible outcomes for only the first two tosses 2*2=4

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

For first toss is heads, second tails and third heads there is only one possible way for you to get this outcome so there is only one element to this set.

{HTH}

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