find the quadratic equation whose roots are 3/2 and5/2? is there anyoe who could please help me?
(x - 3/2)(x - 5/2)
do you follow? you can expand it to get it in standard form
i understand so far
A root is just what makes the equation zero so you just need to make an equation where the equation is rendered zero when you input those variables as x
If you have an equation whose roots are \(r_1, r_2, r_3, ...\) it can always be written as \((x-r_1)(x-r_2)(x-r_3)....\) Just multiply it out.
its pretty evident how to do it once you see it
yes
actually, it should be \(k(x-r_1)(x-r_2)...\) as the multiplication by a constant value \(k\) does not affect the roots.
??
you start with (x - 3/2)(x - 5/2) =0 but you will get ugly fractions that way. if you multiply both by 2: 2(x-3/2) * 2 * (x- 5/2) (2x -3) * (2x-5) now multiply it out
so it will be 2x^2 -6x-15??
Use FOIL to multiply it out. F (first), Outer, Inner, Last
(2x -3) (2x-5) the "First terms" are 2x and 2x. what is 2x*2x ?
4x right?
when you multiply 2*x*2*x you can change the order of the multiplies, so for example, 2*2*x*x
2*2 is 4 but x*x is not just x
4x^2
yes, the answer is 2*2*x*x, but people simplify it to 4*x*x and then they use the "short-hand" x^2 for x*x (just to make it look easier) so you have 4x^2 next, do Outer the outer terms are the very first term and the very last term in (2x -3) (2x-5) ^ ^
would it be -10x 0r just 10 x
it is 2x* -5 = -10x (the minus sign "sticks" to the number or letter) to remember this, always use addition, and write (2x - 5) as (2x + -5) now it is more clear you should use -5
next do Inner (2x -3) (2x-5) ^ ^
-6x
next, Last (2x -3) (2x-5) ^ ^
15
so we have 4 products: 4x^2 -10x -6x + 15 you can combine the x terms (the 2 middle terms) think of it as -10 x's - 6 x's (confusing with the minus signs) but you get -16 x
so it will be 4x^2- 16x+15?
the answer is y = 4x^2 -16x + 15 as a check, see http://www.wolframalpha.com/input/?i=solve+0+%3D+4x%5E2+-16x+%2B+15
thank you sooooo muc
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