Math, asked by angel614, 6 hours ago

show your solutions follow the steps and then answer​

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Answered by tennetiraj86
5

Answer:

17) ( d)

18) (c)

19) (b)

20) ( c)

Step-by-step explanation:

17) f(x) = 3x²-2x+1 has no real solutions

Given function is f(x) =3x²-2x+1

On comparing with ax²+bx+c

a = 3

b = -2

c = 1

b²-4ac = (-2)²-4(3)(1)

=> b²-4ac = 4-12

=> b²-4ac = -8

=> b²-4ac < 0

We know that

If ax²+bx+c has no real solutions then b²-4ac < 0

f(x) = 3x²-2x+1 has no real solutions.

18) k = 3

Given equation is f(t) = t²+kt+3

If k = 3 then f(t) = t²+3t+3

On comparing with ax²+bx+c

a = 1

b = 3

c = 3

b²-4ac = 3²-4(1)(3)

=> b²-4ac = 9-12

=> b²-4ac = -3

=> b²-4ac < 0

We know that

If ax²+bx+c has no real solutions then b²-4ac < 0

19) 36 should be added

Given that x²+12x = 13

=> x²+2(x)(6) = 13

On adding 6² = 36 both sides then

=> x²+2(x)(6)+36 = 13+36

=> (x+6)² = 49

=> x+6 = ±√49

=> x+6 = ±7

=> x = ±7-6

=> x = 7-6 or -7-6

=> x = 1 or -13

The solutions are 1 and -13

20)

4x²+2x-2 ≥ 0 is the Quadratic inequality.

Because it is of the degree 2 and having inequation.

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