I need help i haven't had this sense last year and i need help understanding how this question works z^2+z-6=0

First factor it:


(z+3)(z-2)=0





Then you have two solutions:


z=-3 and z=2





Substituting either causes the product (z+3)(z-2) to be 0.I need help i haven't had this sense last year and i need help understanding how this question works z^2+z-6=0
Simplest way to solve quad equations is by using quad formula:





for ax2 + bx + c = 0


then x = (-b +/- SQRT(b^2-4ac))/2a





in your case


x = (-1 +/- SQRT(1^2-4(1)(-6))/2(1)





so x = 2 or x = -3I need help i haven't had this sense last year and i need help understanding how this question works z^2+z-6=0
Hi there,


I think all of the answers they've given you doesnt really show how to factor them so here's a tip.


Your equation is in the form ax^2 + b + c.


and


Ax^2 + B + C can be solved by the following:





because the degree of this equation (highest power) is 2


we have


( x _____ )( x _____)


A determines the product of the coefficients of x





in your equation, A=1 so we have (x ___ )(x ___ )





the sign of C determines the signs of what goes inside the brackets


if we have a positive (+)


we could either have (x + __ )(x +__ ) or (x - __ )(x -__ )


if we have negative (-)


we have, (x + __ )(x -__ )





on your equation, we have a negative so we have


(x + __ )(x -__ ) ===%26gt; (*)





now, think of 2 numbers that when you multiply they would equal to C





on our equation, we have 6


so we have 1 %26amp; 6 or 2 %26amp; 3


now, try substituting them on our equation (*)


make sure that when you add the x parts, you'll have the Bx part


we could not use 1%26amp; 6 as there's no way we could have 1x if we add them, so we'll use 2%26amp;3


(x + __ )(x -__ ) =?(x + 2 )(x - 3)


check: 2x -3x = -x


so we got it wrong, our Bx part should be equal to 1x


(x + __ )(x -__ ) =?(x + 3 )(x - 2)


check: 3x -2x = 1x and it is right





now, we have (x + 3 )(x - 2)=0


now, the question is, how can we have 0 given 2 numbers being multiplied together?


since we know that anything multiplied to 0 is 0,





then it is either the (x+3) part should be zero or the (x-2) should be zero





1. x+3 =0 ==%26gt; x=-3


2. x-2 = 0 ==%26gt; x= 2





cheers


ps. your equation is in terms of z and mine is on x, im sorry
z^2 + z - 6=0


(z+3)(z-2)=0 ---- (2)


z=-3, z=2 --------(3)





cross-multiplication. to get (2), you need to be able to find out how the numbers can be multiplied to get the middle thing. for (3) to work, anything multiplied by zero, one of the factors must equal to zero. hence, the individual brackets must equal zero.
factor out the equation:


z^2+z-6= 0





the roots are (z+3)(z-2)





equate the roots to zero


*z+3=0


z=-3





*z-2=0


z=2





substituting z to the quadratic equation you would get 0
x=2 or -3

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