Math, asked by areebakhan1716, 1 month ago

3q^2=2q+8 use qaudratic formula and solve




please solve this in a book and give me the answer​

Answers

Answered by XxitsmrseenuxX
6

Answer:

{\huge{\pink{↬}}} \:  \: {\huge{\underline{\boxed{\bf{\pink{Answer}}}}}}

How to solve your problem

32=2+8

3q^{2}=2q+83q2=2q+8

Quadratic formula

Factor

1

Move terms to the left side

32=2+8

3q^{2}=2q+83q2=2q+8

32−(2+8)=0

3q^{2}-\left(2q+8\right)=03q2−(2q+8)=0

2

Distribute

32−(2+8)=0

3q^{2}-\left(2q+8\right)=03q2−(2q+8)=0

32−2−8=0

3q^{2}-2q-8=03q2−2q−8=0

3

Use the quadratic formula

=−±2−4√2

q=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}q=2a−b±b2−4ac

Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

32−2−8=0

3q^{2}-2q-8=03q2−2q−8=0

=3

a={\color{#c92786}{3}}a=3

=−2

b={\color{#e8710a}{-2}}b=−2

=−8

c={\color{#129eaf}{-8}}c=−8

=−(−2)±(−2)2−4⋅3(−8)√2⋅3

q=\frac{-({\color{#e8710a}{-2}}) \pm \sqrt{({\color{#e8710a}{-2}})^{2}-4 \cdot {\color{#c92786}{3}}({\color{#129eaf}{-8}})}}{2 \cdot {\color{#c92786}{3}}}q=2⋅3−(−2)±(−2)2−4⋅3(−8)

4

Simplify

Evaluate the exponent

Multiply the numbers

Add the numbers

Evaluate the square root

Answered by munibaabdullah10
0

Answer:

Solution

=

2

=

4

3

Step-by-step explanation:

1

Move terms to the left side

3

2x

=

2

+

8

3q^{2}=2q+8

3q2=2q+8

3

2

(

2

+

8

)

=

0

3q^{2}-\left(2q+8\right)=0

3q2−(2q+8)=0

2

Distribute

3

Use the quadratic formula

4

Simplify

5

Separate the equations

6

and you can refer to this too :-

Correct :-

q=2,

3

−4

Given equation is,

3q

2

=2q+8

3q

2

−2q−8=0

3q

2

−6q+4q−8=0

3q(q−2)+4(q−2)=8

(3q+4)(q−2)=0

∴q=2,

3

−4

∴ roots of the given equation are 2,

3

−4

mark as brainliest answer

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