Question

4x²+12x+5 find zeroes​

Answer 1

Answer:

a=4,b=12 & c=5

D=b^2 - 4ac

=(12)^2 - 4*4*5

= 144 - 80

= 64

Let the zeroes be α & β.

α= -b-√D/2a

= -12-√64/2*4

= -12-8/8

= -20/8

= -5/2

β=-b+√D/2a

= -12+√64/2*4

= -12+8/8

= -4/8

= -1/2

Therefore the zeroes are -5/2 & -1/2.

Answer 2

Answer:

$zeroes = \frac{ - 1}{2} and \frac{ - 5}{2}$

Step-by-step explanation:

4x² + 12x + 5

In the standard form of quadratic equation (ax² + bx + c), here

a = 4

b = 12

c = 5

.

To find the zeroes, we will use the zeroes formula

$zeroes \: = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ = \frac{ - 12 ± \sqrt{ {12}^{2} - 4(4)(5) } }{2(4)} \\ \\ = \frac{ - 12 ± \sqrt{144 - 80} }{8} \\ \\ = \frac{ - 12 ± \sqrt{64} }{8} \\ \\ = \frac{ - 12 ± 8}{8} \\ (as + ) \\ = \frac{ - 12 + 8}{8} \\ = \frac{ - 4}{8} \\ = \frac{ - 1}{2} \\ (as - ) \\ \frac{ - 12 - 8}{8} \\ = \frac{ - 20}{8} \\ = \frac{ - 5}{2}$

Hence zeroes = -1/2 and -5/2

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hope it helps.