Hands first, words secondDrag the two roots on the number line to reach the mini goal. Then move only the q slider, making it bigger and bigger. What happens to the two roots?
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Try this
Drag only the right-hand root 1 step to the right. How much does the sum of the roots change? How much does p change? In the same direction?
Drag one root to 0. What is the product of the roots? And the constant term q?
Leave the roots alone and move only the q slider. The midpoint (the small blue triangle) stays put while the two roots move toward each other or apart. If the sum stays the same, what happens to the product as the roots get closer?
Keep making q bigger until the two roots collide. What is Δ now? Go one step further: where did the roots go?
Predict first
Without solving: x1 and x2 are the roots of x2−7x+3=0. What is x12+x22?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
The roots are 27±37, and squaring them and adding is messy. Try another way:
x12+x22=(x1+x2)2−2x1x2=72−2×3=43.
The sum 7 and the product 3 can be read straight off the coefficients. You never need to know the two roots themselves.
(Do not forget to glance at Δ=49−12=37>0 first: the two roots really exist.)
Why
Why are the sum and product of the roots hidden in the coefficients? Because once you know the two roots, you can “multiply out” the equation.
From two roots to two coefficients
Step 1 / 6
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Roots first, then the equation
We want an equation whose roots are exactly 2 and 5: just set the product of x−2 and x−5 equal to 0.
(x−2)(x−5)=0
Multiply out and look at the linear term
Now the constant term
Any pair of roots works the same way
A general quadratic equation
Use it backwards
Symmetric expressions without solving
If swapping x1 and x2 leaves an expression unchanged, it is called a symmetric expression. Every one of them can be built from the sum and the product:
Below is x2−4x+1=0. Its roots, 2±3, contain square roots, yet the sum of their squares is a clean integer. Drag the roots and watch how these two symmetric expressions change.
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Notice that (x1−x2)2 always equals Δ: the distance between the roots is Δ.
Remove one condition
Break it
Counterexample lab: remove one condition
Conclusion: you can calculate with x1+x2=−p and x1x2=q← still true?
x2−5x+6=0: Δ=1>0 and the roots are 2 and 3. Sum of squares =52−2×6=13, the same as working out 4+9 directly.
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Skip the check on Δ and apply the formula to x2+2x+3=0: “x12+x22=(−2)2−2×3=−2”.
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Two real numbers whose squares add up to a negative number? Impossible. The picture shows why: Δ=−8<0, so the two roots have left the number line.
There are no real x1 and x2, so “their sum” and “their sum of squares” mean nothing. Move q down
and watch for the moment the sum of squares starts to make sense.
Practice
Exercise 1Basic
Without solving, write down the sum and the product of the roots of 2x2−6x−5=0. (Separate the two numbers with a comma.)
Hint · direction
The leading coefficient is not 1, so reading the coefficients directly will go wrong.
Hint · method
x1+x2=−ab and x1x2=ac.
Hint · first step
a=2,b=−6,c=−5.
Full solution
x1+x2=−2−6=3 and x1x2=2−5=−25. (Δ=36+40>0, so the roots exist.)
Exercise 2Basic
One root of x2+kx−12=0 is 3. Without finding k, find the other root.
Hint · direction
Which of the two relations does not involve k?
Hint · method
The product of the roots equals the constant term.
Hint · first step
3⋅x2=−12.
Full solution
x2=−4. (And by the way: 3+(−4)=−1=−k, so k=1.)
Exercise 3Advanced
x1 and x2 are the roots of x2−4x+1=0. Find x2x1+x1x2.
Hint · direction
Put it over a common denominator first. Which symmetric expressions are the numerator and the denominator?
Hint · method
x2x1+x1x2=x1x2x12+x22. Then write the numerator in terms of the sum and the product.
Hint · first step
x1+x2=4 and x1x2=1, so x12+x22=16−2=14.
Full solution
x1x2x12+x22=114=14. (Δ=12>0, so the roots exist and neither is 0.)
Exercise 4Challenge
The equation x2−2mx+2m+3=0 in x has two real roots x1 and x2. Find the minimum value of x12+x22.
Hint · direction
“Has two real roots” is not just politeness: it limits the values m can take.
Hint · method
First use Δ≥0 to find the range of m. Then write x12+x22 in terms of m and find its minimum over that range.
Hint · first step
Δ=4m2−8m−12≥0, that is, (m−3)(m+1)≥0, so m≤−1 or m≥3.
Full solution
x12+x22=(2m)2−2(2m+3)=4m2−4m−6=4(m−21)2−7. But m=21 is not in the allowed range. The allowed values closest to 21 are m=−1 (distance 1.5) and m=3 (distance 2.5), so the minimum is at m=−1: 4+4−6=2. The equation is then (x+1)2=0, both roots are −1, and the sum of squares really is 2.
The wrong step: For x2−5x+6=0 the sum of the roots is −5 and the product is 6.
What went wrong? Think first, then open
DiagnosisExpand (x−x1)(x−x2) and the linear term is −(x1+x2)x, so the sum of the roots is the opposite of the linear coefficient. Here the roots are 2 and 3, and their sum is clearly +5.
✓
The right step: x1+x2=−ab=−1−5=5 and x1x2=ac=6.
“The sum flips sign, the product does not”
Exhibit
Using Vieta's formulas without checking Δ
✗
The wrong step: The equation x2−2mx+m+2=0 has two real roots x1,x2. x12+x22=(2m)2−2(m+2)=4(m−41)2−417, so the minimum is −417.
What went wrong? Think first, then open
DiagnosisA sum of squares of two real numbers can never be negative! The trouble is that at m=41 we get Δ=4m2−4m−8<0: the equation has no real roots at all, so x1,x2 do not exist.
✓
The right step: First make sure real roots exist: Δ≥0 means m2−m−2≥0, so m≤−1 or m≥2. In that range 4m2−2m−4 is smallest at m=−1, where it equals 2 (and both roots are −1).
“Vieta's formulas are about roots, so first make sure the roots exist”
Exhibit
Forgetting to divide by a when the leading coefficient is not 1
✗
The wrong step: For 2x2−6x−5=0 the sum of the roots is 6 and the product is −5.
What went wrong? Think first, then open
DiagnosisYou read the sum and product straight off the coefficients only for an equation like x2+px+q=0, with leading coefficient 1. Here the leading coefficient is 2, so divide both sides by 2 first.
✓
The right step: Divide by 2: x2−3x−25=0. So x1+x2=3 and x1x2=−25.
“Make the leading coefficient 1 first, or remember to divide by a”
Vieta's Formulas ← → step · F full screen · Esc exit