Maths Olympiad Prep

Track / Stage 4 / 247 of 340 #507 of 1964

Problem 507

AMC 12 late, AIME early
Algebra Difficulty 4.9 Find the answer

6.066. x2−4x−6=2x2−8x+12x^{2}-4 x-6=\sqrt{2 x^{2}-8 x+12}.

A number or a short expression. Spacing and $ signs are ignored.

Official solution

Solution.

By squaring both sides of the equation, we have

(x2−4x−6)2=2x2−8x+12⇔(x2−4x−6)2−2(x2−4x−6+12)=0 \left(x^{2}-4 x-6\right)^{2}=2 x^{2}-8 x+12 \Leftrightarrow\left(x^{2}-4 x-6\right)^{2}-2\left(x^{2}-4 x-6+12\right)=0

Let x2−4x−6=y,y≥0x^{2}-4 x-6=y, y \geq 0. The equation in terms of yy becomes y2−2y−24=0y^{2}-2 y-24=0, from which y1=−4,y2=6;y1=−4y_{1}=-4, y_{2}=6 ; y_{1}=-4 is not valid.

Then x2−4x−6=6⇔x2−4x−12=0,x1=−2,x2=6x^{2}-4 x-6=6 \Leftrightarrow x^{2}-4 x-12=0, x_{1}=-2, x_{2}=6. By verification, we confirm that these are indeed the roots of the original equation.

Answer: x1=−2,x2=6x_{1}=-2, x_{2}=6.

Solve the systems of equations (6.067-6.119):

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.