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uranmaximum [27]
3 years ago
14

*HELP ME ASAP I NEED THIS AWNSER*

Mathematics
1 answer:
SashulF [63]3 years ago
5 0

Answer: 1120 feet

Step-by-step explanation:

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Find the zeros the polynomial fx=2x^2 +9x-5<br>pls help with steps ​
lara [203]

Answer:

-5 ; 1/2

Step-by-step explanation:

<u>GIVEN :-</u>

  • A quadratic polynomial f(x) = 2x² + 9x - 5

<u>TO FIND :-</u>

  • Zeroes of f(x) = 2x² + 9x - 5

<u>GENERAL CONCEPTS TO BE USED IN THIS QUESTION :-</u>

Lets say there's a quadratic polynomial f(x) = ax² + bx + c , whose factors are (x - α) & (x - β). To find the values of x for which f(x) will be zero , equate the factors of f(x) with 0.

⇒ (x - α) = 0 & (x - β) = 0

⇒ x = α & x = β

Hence, it can be concluded that if (x - α) & (x - β) are factors of f(x) , then α & β are the roots of f(x).

<u>SOLUTION :-</u>

Factorise f(x) = 2x² + 9x - 5.

  • Split its middle term.

=> 2x^2 + 10x - x - 5

  • Take '2x' common from first two terms & '-1' from last two terms.

=> 2x(x + 5) - 1(x + 5)

  • Take (x + 5) common from the whole expression.

=> (x + 5)(2x - 1)

So , the factors of f(x) are (x + 5) & (2x - 1). Now equate the factors with zero.

⇒ (x + 5) = 0 & (2x - 1) = 0

⇒ x = -5 & x = 1/2

∴ The zeroes of f(x) = 2x² + 9x - 5 are (-5) & (1/2)

<u>VERIFICATION :-</u>

1) Put x = -5 in f(x) = 2x² + 9x - 5

⇒ f(-5) = 2(-5)² + 9×(-5) - 5

           = 50 - 45 - 5

           = 50 - 50  

           = 0

2) Put x = 1/2 in f(x) = 2x² + 9x - 5

f(\frac{1}{2} ) = 2 \times (\frac{1}{2} )^2 + 9 \times \frac{1}{2}  - 5

       = 2 \times \frac{1}{4} + \frac{9}{2} - 5

       = \frac{1}{2} + \frac{9}{2} - 5

       = \frac{ 1 + 9 - 10}{2}

       = \frac{0}{2}

       = 0

6 0
3 years ago
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