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2 years ago

15

Solve the quadratic equation 3x^2 +x-5=0 give your answers to two decimal places

1 answer:

ddd [48]2 years ago

8 0

Especially if we're to present the answers to 2 decimal places, we'd gues that the roots of this polynomial are not "nice."

My suggestion is that we use the quadratic formula to find the roots, and then write the associated factors.

In 3x^2 +x-5=0, a=3, b=1 and c= -5. Thus, the roots are

-1 plus or minus √(1^2 - 4(3)(-5) )

x = --------------------------------------------------

2(3)

or:

-1 plus or minus √61 -1 plus or minus 7.81

x = ----------------------------------- = ---------------------------------

6 6

Then one root is 6.81/6, or approx. 1.14, and the other is

-8.81/6, or approx. -1.47.

The solution set is then {-1.47, 1.14}.

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The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
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Step-by-step explanation:

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$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

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The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

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for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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