Want to Ace Rational Equations? Watch This Now!

[@alol6956]

Wow, you really went much faster in your steps in this video. It was so surprising

[@Shobhamaths]

x=(7±√47)/2,(9±√67)/2 a=(4x^2-30x+5)/(2x-3) a^2+(a-2)^2=10 a^2-2a-3=0 a==-1,3 Simplify by putting a value we get x

[@RashmiRay-c1y]

Let [(4x^2-30x+5)/(2x-3)]=a and [4x^2-34x+11)/(2x-3)]=b. Then, a^2+b^2=10 and a-b=2. Thus, a=-1,2. If a=-1, 2x^2-14x+1=0 and x=1/2[7+/-√47]. If a=3, 2x^2-18x+7=0 and x = 1/2[9+/-√67]. So, x=1/2[7+/-√47], 1/2[9+/-√67].

[@gregevgeni1864]

Obviously 2x -3 no zero, x=/3/2 . 4x²-30x+5=4x²-12x+9-18x-4=(2x-3)²-9(2x-3)-27-4=(2x-3)²-9(2x-3)-31 => (4x²-30x+5)/(2x-3) = [(2x-3)²-31-9(2x-3)]/(2x-3)=(2x-3)-31/(2x-3) -9 =M - 9 (1). with M =(2x-3)-31/(2x-3) (*) Likewise 4x²-34x+11=(2x-3)²-11(2x-3)-31 => (4x²-34x+11)/(2x-3) = (2x-3)-31/(2x-3)-11=M -11 (2). From (1), (2) the given equation is equivalent to (M - 9)² +(M -11)² = 10 or M² -20M +192 = 0 => M = 8 or M = 12 (3). From (*) and (3) => (2x-3)-31/(2x-3) = 8 => x = (7±sqrt (47))/2. Too, (2x-3)-31/(2x-3) = 12 => x = (9±sqrt (67))/2.

[@ShriH-d1o]

Let a²+b²=10; & solving a - b=2,; >a=b+2; =>(b+2)²+b²=10; 2b²+4b-6=0;>(b+3)(b-1) =0; b=1 or -3; a=3 or -1; For b=1;=>2x²-18x+7=0; x=(9±√67)/2 ... (1) for b=-3; >2x²-14x+1=0; => x=(7±√47)/2 .. (2)

[@RealQinnMalloryu4]

({16x^4 ➖ 900x^2}+10/4x^2 ➖) 9)+({16x^4 ➖ 1156x^2}22/4x^2 ➖ 9)=({886x^2+10}/5x^2)+({1140x^2+22}/5x^2{ 896x^2/5x^2+1162x^2/5x^2}=2058x^4/10x^4=258x^1 10^20^2^29^1x^1 1^2^2^1^1x^1 1^2^x (x ➖2 x+1).

[@Quest3669]

X=(9+-√67)/2; (7+-√47)/2 solns.

[@潘博宇-k4l]

X No 2/3, X1=(9+√67)/2, X2=(9-√67)/2, X3=(7+√47)/2, X4=(7-√47)/2.