5x2+7x-3=0 wats te discriminant
huh
so how do u find if it is a real or imaginary
\[ ax^2+bx+c\]\[Discriminant = \,b^2-4ac. \]
so how do i find it with tthat only and how is it real or imaginary
well in your question a = 5, b = 7 and c = -3 substitute and evaluate
oh so the a will be 5
put u dont have to take out the five and put it into a diffrent equation
not quite... I think you need to try again...
oh ok thank u that was really helpful bro
wait so whats real and imaginary mean
ok ithe calculation is \[7^2 - 4 \times 5 \times (-3) =\] if its greater than zero it has real roots if its equal to zero it has 1 repeated root is it less than zero then you complex roots...
oh i get it thanks u mind when i need help i can ask u bro
thats ok... if I can do it
lmao thanks bro
i have another quetsion
ok
ok i solved another question and i got 0 but it says to write if it is a real number tell weather they will be ration or irrational how would i do that im confused
ok.... so if you get a zero value the quadratic is a perfect square... whats the equation...?
so would that be a real number so is it rational or irrational
here is an example y=x^2 + 6x + 9 or y = (x + 3)^2 has a repeated root at x = -3 and the discriminant is 36 - 36 = 0
ok so is it rational or irrational
so a parabola such as \[y=x^2 - 6x + 5 ....= (x -5)(x -1) \] has roots at 1, 5 the discriminant 36 - 20 = 16 16 is a perfect square so the parabola has 2 unequal roots that are real and rational.... if you think about your original question the discriminant was 109.... which means 2 unequal real roots which are irrational as 109 isn't a square number.
oh ok so an unperfect square is a irrational and a perfect square would be rational
ok so x2 would be a=1 right
yes... as long as its >0
ok y that face
good luck
lmao thanks ur great help bro
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