ABCD is a rectangle in which DP and BQ are perpendicular from D and B respectively on the diagonal AC. Prove that DP=BQ.
Answers
Step-by-step explanation:
1
2
If the roots of ax2+bx+c=0 are equal then the equal root is what?
Galaxy S21 | S21+ 5G.
Dean Rubine
Answered July 8, 2020
There are a few ways to see this one.
Viete’s formula tells us the sum of the roots is −b/a, so if the roots are the same, they’re both half that:
x=−b2a
Let’s see where the sum of the roots comes from. If we knew the roots r and s we could factor our quadratic as
ax2+bx+c=a(x−r)(x−s)=ax2−a(r+s)x+ars
so −a(r+s)=b , that is the sum of the roots r+s=−b/a.
We know for a single root our quadratic factors as a perfect square:
ax2+bx+c=a(x−r)2=ax2−2arx+r2
so equating respective coefficients
b=−2ar,c=ar2
and we again get
r=−b2a
precisely when c=ar2=b24a , i.e. the zero discriminant:
b2−4ac=0
The quadratic formula is of course
x=−b±b2−4ac−−−−−−−√2a
The roots are equal when we have ±0 , so a zero radicand:
b2−4ac=0
which tells us
x=−b2a
Equal roots means the vertex of the parabola is the zero — the parabola is tangent to the x axis. So the zero is the minimum or maximum, which we can find by completing the square:
ax2+bx+c=a(x2+bax)+c=a(x2+ba+b24a2)+b24a+c=a(x+b2a)2+b24a+c
That clearly has a min or max (which depends on the sign of a ) at
x=−b2a
and that’s a zero precisely when
b24a+c=0
which is again the zero discriminant.
In the quadratic
x2−2Bx+c=0
the sum of the roots is 2B so their average is B. If the two roots are the same, they’ll be B .
We get a single root precisely when
B2−2B(B)+c=0
c=B2
which is again the zero discriminant for this equation.
That was pretty boring. Let’s do something slightly more interesting to end, and solve simultaneous quadratic equations:
ax2+bx+c=0
dx2+ex+f=0
These equations may have zero, one or two roots in common.
Our first step is to eliminate the quadratic terms.
adx2+bdx+cd=0
adx2+aex+af=0
Subtracting,
(ae−bd)x+(af−cd)=0
x=−af−cdae−bd
Our system is overdetermined so this might not be a solution at all. Let’s solve again by eliminating linear terms:
aex2+bex+ce=0
bdx2+bex+bf=0
(ae−bd)x2+(ce−bf)=0
x2=bf−ceae−bd
We’ll only get a solution when the two ways of calculating x2 are the same,
(−af−cdae−bd)2=bf−ceae−bd
When we clear the fractions we get an expression that tells us if we have solutions:
(af−cd)2=(bf−ce)(ae−bd)
OK, now we have a handle on simultaneous quadratic equations. Let’s sum up.
The Simultaneous Quadratic Formula.
Solutions exist for the system
ax2+bx+c=0,dx2+ex+f=0a≠0,d≠0,
precisely when (af−cd)2=(bf−ce)(ae−bd).
If so, when ae≠bd we have exactly one solution,
x=−af−cdae−bd
Otherwise ae=bd so af=cd and our two quadratic equations are equivalent, with solutions given by the usual quadratic formula
x=−b±b2−4ac−−−−−−−√2a