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Chapter 9: Q. 9.42 (page 889)

Solve by using the Quadratic Formula:4z2+2z-6=0

Short Answer

Expert verified

The roots of the above equation isz=1andz=-32

Step by step solution

01

Given information

We are given a quadratic equation4z2+2z-6=0

02

Step 2: Identify the values of a,b,c

On comparing with standard form we get the values ofa,b,casa=4,b=2,c=-6

03

Write the quadratic equation formula and substitute the values of a,b,c and then simplify and solve for x

z=-b卤b2-4ac2a

Substituting the values we get

z=-2卤4-4(4)(-6)2(4)z=-2卤4+968z=-2卤1008z=-2卤108

On further simplifying we get

z=-2+108z=88z=1andz=-2-108z=-128z=-32

04

Step 4:  check the solutions

We substitute x=1in the quadratic equation

4z2+2z-64(1)2+2(1)-64+2-6=0

Now we substitute z=-32in the above equation

4z2+2z-64(-32)2+2(-32)-69-3-6=0

05

Conclusion

The solution of the above equation isz=1andz=-32

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