Easy and Simple

Algebra Level 2

The equation 2 x 2 + q x + ( q − 1 ) = 0 2x^2 + qx + (q-1) = 0 has the roots x x and y y . If x 2 + y 2 = 4 x^2 + y^2 = 4 , find q q

− 6 -6 and 2 2 − 3 -3 and 5 5 − 6 -6 and − 2 -2 − 4 -4 and 4 4 − 2 -2 and 6 6

Jason Chrysoprase by Jason Chrysoprase , 18, Indonesia

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2 solutions

By Vieta's Formula.

→ x + y = − q 2 \rightarrow x+y=\frac{-q}{2}
→ x y = q − 1 2 \rightarrow xy=\frac{q-1}{2}
Now,
⇒ x 2 + y 2 = ( x + y ) 2 − 2 x y = 4 \Rightarrow x^2+y^2=(x+y)^2-2xy=4
⇒ ( − q 2 ) 2 − 2 × ( q − 1 2 ) = 4 \Rightarrow \left(\frac{-q}{2}\right)^2-2× \left(\frac{q-1}{2}\right)=4
⇒ q 2 − 4 q − 12 = 0 \Rightarrow q^2-4q-12=0
⇒ ( q + 2 ) ( q − 6 ) = 0 \Rightarrow (q+2)(q-6)=0
∴ q = 6 , − 2 \therefore q=6,-2

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x 2 + y 2 = 4 x^2 + y^2 = 4

( x + y ) 2 − 2 x y = 4 (x+y)^2 - 2xy = 4

By using Vieta's Formula

( − q 2 ) 2 − 2 ( q − 1 2 ) = 4 (\frac{-q}{2})^2 - 2(\frac{q-1}{2}) = 4

q 2 4 − q + 1 = 4 \frac{q^2}{4} - q + 1 = 4

q 2 − 4 q + 4 = 16 q^2 - 4q +4 = 16

q 2 − 4 q − 12 = 0 q^2 - 4q - 12 = 0

( q − 6 ) ( q + 2 ) = 0 (q-6)(q+2)= 0

q = 6 q = 6 or q = − 2 q = -2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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