Here is the solution to Question 9.
Question 9:
Given the quadratic equation 3x2−11x+4=0, with roots α and β.
From the properties of quadratic equations, for ax2+bx+c=0:
Sum of roots: α+β=−ab
Product of roots: αβ=ac
For 3x2−11x+4=0:
α+β=−3(−11)=311
αβ=34
We need to find the equation whose roots are X1=α2+β1 and X2=α1+β2.
A quadratic equation with roots X1 and X2 is given by x2−(X1+X2)x+(X1X2)=0.
Step 1: Calculate the sum of the new roots, S=X1+X2.
S=(α2+β1)+(α1+β2)
S=α2+β2+α1+β1
We know that α2+β2=(α+β)2−2αβ.
And α1+β1=αβα+β.
First, calculate α2+β2:
α2+β2=(311)2−2(34)
α2+β2=9121−38
α2+β2=9121−924=997
Next, calculate α1+β1:
α1+β1=4/311/3=411
Now, substitute these values back into the expression for S:
S=997+411
To add these fractions, find a common denominator, which is 36.
S=9×497×4+4×911×9
S=36388+3699=36487
Step 2: Calculate the product of the new roots, P=X1X2.
P=(α2+β1)(α1+β2)
Expand the product:
P=α2⋅α1+α2β2+β1⋅α1+β1β2
P=α+(αβ)2+αβ1+β
Rearrange the terms:
P=(α+β)+(αβ)2+αβ1
Substitute the values of α+β and αβ:
P=311+(34)2+4/31
P=311+916+43
To add these fractions, find a common denominator, which is 36.
P=3×1211×12+9×416×4+4×93×9
P=36132+3664+3627
P=36132+64+27=36223
Step 3: Form the new quadratic equation x2−Sx+P=0.
Substitute the calculated values of S and P:
x2−36487x+36223=0
Multiply the entire equation by 36 to eliminate the denominators:
36x2−487x+223=0
36x2−487x+223=0
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