Question

1) 2y^2+y-10=0 2) 3p^2-10p+3=0 3) 5v^2-9v-2=0 4) 12z^2+5z-2=0 5) 17s^2-19s+2=0

Answer 1

......................

Answer 2

1) 2y^2+y-10=0
2y^2+5y-4y-10=0
y(2y+5)-2(2y+5)=0
(2y+5)(y-2)=0
2y+5=0
y-2=0
y1= -5/2
y2=2

2) 3p^2-10p+3=0
3p^2-p-9p+3=0
p(3p-1)-3(3p-1)=0
(3p-1)(p-3)
3p-1=0
p-3=0
p1=1/3
p2=3

3)5v^2-9v-2=0
5v^2+v-10v-2=0
v(5v+1)-2(5v+1)=0
(5v+1)(v-2)=0
5v+1=0
v-2=0
v1= -1/5
v2=2

4)12z^2+5z-2=0
12z^2+8z-3z-2=0
4z(3z+2)-(3z+2)=0
(3z+2)(4z-1)=0
3z+2=0
4z-1=0
z1= -2/3
z2=1/4

5)17s^2-19s+2=0
17s^2-2s-17s+2=0
s(17s-2)-(17s-2)=0
(17s-2)(s-1)=0
17s-2=0
s-1=0
s1=2/17
s2=1

Answer 3

To solve the quadratic equations you provided, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's solve each equation step by step:

1. 2y^2 + y - 10 = 0

a = 2, b = 1, c = -10

x = (-(1) ± √(1^2 - 4 * 2 * (-10))) / 2 * 2 x = (-1 ± √(1 + 80)) / 4 x = (-1 ± √81) / 4 x = (-1 ± 9) / 4

So the two solutions for y are: y = (8 / 4) = 2 y = (-10 / 4) = -2.5

2. 3p^2 - 10p + 3 = 0

a = 3, b = -10, c = 3

x = (-(10) ± √((-10)^2 - 4 * 3 * 3)) / 2 * 3 x = (10 ± √(100 - 36)) / 6 x = (10 ± √64) / 6 x = (10 ± 8) / 6

So the two solutions for p are: p = (18 / 6) = 3 p = (2 / 6) = 1/3

3. 5v^2 - 9v - 2 = 0

a = 5, b = -9, c = -2

x = (-(9) ± √((-9)^2 - 4 * 5 * (-2))) / 2 * 5 x = (9 ± √(81 + 40)) / 10 x = (9 ± √121) / 10 x = (9 ± 11) / 10

So the two solutions for v are: v = (20 / 10) = 2 v = (2 / 10) = 0.2

4. 12z^2 + 5z - 2 = 0

a = 12, b = 5, c = -2

x = (-(5) ± √(5^2 - 4 * 12 * (-2))) / 2 * 12 x = (-5 ± √(25 + 96)) / 24 x = (-5 ± √121) / 24 x = (-5 ± 11) / 24

So the two solutions for z are: z = (6 / 24) = 0.25 z = (-16 / 24) = -0.67

5. 17s^2 - 19s + 2 = 0

a = 17, b = -19, c = 2

x = (-(19) ± √((-19)^2 - 4 * 17 * 2)) / 2 * 17 x = (19 ± √(361 - 136)) / 34 x = (19 ± √225) / 34 x = (19 ± 15) / 34

So the two solutions for s are: s = (34 / 34) = 1 s = (4 / 34) = 2/17

These are the solutions for the given quadratic equations.