What is the value of x in the equation square root of (5-2x) = 3i?
√(-2x + 5) = 3i; square both sides √(-2x + 5)^2 = (3i)^2; simplify -2x + 5 = (3i)(3i); simplify the right side -2x + 5 = 9i^2; i^2 = -1 -2x + 5 = 9(-1); simplify -2x + 5 = -9; subtract 5 from both sides -2x + 5 - 5 = -9 - 5; simplify -2x = -14; divide both sides by -2 -2x/-2 = -14/-2; solve x = 7 <-----------Answer Blessings
What is the domain of g(x)=square root of (2x-9)?
The domain is the set of all number you can plug into a function. You can't take the square root of a negative number, so we're OK so long as 2x - 9 is greater than or equal to 0. So the domain is the set of all real values of x where x >= 9/2.
What is square root of 4x-3= 2+square root of 2x-5?
Hi, . _____ . . . . . _____ √4x - 3 = 2 + √2x - 5 Square both sides of the equation. . ._____ . . . . . . ._____ (√4x - 3 )²= (2 + √2x - 5)² . . . . . . . . . . . _____ 4x - 3 = 4 + 4√2x - 5 + 2x - 5 . . . . . . . . _____ 4x - 3 = 4√2x - 5 + 2x - 1 . . . . . . . . _____ 2x - 2 = 4√2x - 5 Divide every term by 2. . . . . . . . _____ x - 1 = 2√2x - 5 Square both sides of the equation. . . . . . . . . . _____ (x - 1)² = (2√2x - 5)² x² - 2x + 1 = 4(2x - 5) x² - 2x + 1 = 8x - 20 x² - 10x + 21 = 0 (x - 7)(x - 3) = 0 x = 7 or x = 3 <==ANSWER Both answers check in original problem. I hope that helps!! :-)
Solve x = square root of 2x - 7 + 5?
Subtract 5 from both sides: x - 5 = (sqrt(2x - 7) Square both sides: (x - 5)² = 2x - 7 Expand out the square: x² - 10x + 25 = 2x - 7 Put everything on one side: x² -12x + 32 = 0 Factor: (x - 4)(x - 8) = 0 Therefore: x = 4 or x = 8 Double-checking: sqrt(2(4) - 7) + 5 sqrt(1) + 5 sqrt(1) is -1 or 1, so x can be 4 <-- check sqrt(2(8) - 7) + 5 sqrt(16 - 7) + 5 sqrt(9) + 5 sqrt(9) is -3 or 3, so x can be 8 <-- check
On what intervals is y= (square root of x^2 +9) continuous?
-infinity to infinity, fairly sure of it
Solve for x : root of 2x + 9 +x= 13?
2x + 9 + x = 13 3x = 4 x = 4/3
If the roots of 2x^2+7x+5=0 are the reciprocal roots of ax^2+bx+c=0, then how do you find the value of a-c?
Let p and q are the roots of 2x^2+7x+5 = 0.p+q = -7/2………….(1)p.q = 5/2……….. ….(2)Thus , 1/p and 1/q are the roots of ax^2+bx+c = 0or x^2 + b/a.x+ c/a = 01/p+1/q = -b/a ………………(3)1/p.1/q = c/a……………………(4).from eq (3)(q+p)/p.q = -b/a(-7/2)/(5/2)= - b/aor b/a=7/5…………..(5)from eq. (4)1/p.q = c/ac/a = 2/5…………………(6)The given eq. is x^2+(b/a).x+c/a =0or x^2+(7/5).x+2/5 =0or 5x^2+ 7x + 2 = 0…………….(7)equating the given eq. ax^2+bx+c=0 to eq.(7)a =5 , b = 7 and c = 2 .And a-c = 5 - 2 =3 , Answer.
If α, β are roots of equation 2x^2-3x-5=0, how would you form a quadratic equation whose roots are α ^2, β^2?
quadratic equation solution is x=(-(-3) pm sqrt((-3)^2–4*2*(-5)))/(2*2)=(3 pm sqrt(9+40))/4=(3 pm 7)/4= {10/4=5/2,-1} so roots getting squared meansthe new poly has roots alp^2=25/4, beta^2=1 so (x-alp^2)*(x-beta^2)=(x-25/4)*(x-1) is the new equation. This can be rationalized to get the numerator as 4*x^2–29*x+25=0. You can also use algebra to realize that the sum of the roots is the negative of the coeff of x, and the product of the roots is the constant term, when the eqn is written as (x-a)*(x-b)=x^2-(a+b)*x+a*b, so x^2–3*x/2–5/2 after dividing by 2 gives a+b=3/2, a*b=-5/2. Thus (a+b)^2–2*a*b=a^2+b^2=(3/2)^2–2*(-5/2)=9/4+5=29/4 and a^2*b^2=(a*b)^2=(-5/2)^2=25/4, so the eqn with roots alp^2 and beta^2 as roots isx^2-(alp^2+beta^2)*x+alp^2*beta^2=x^2–29*x/4+25/4 which again rationalizes to 4*x^2–29*x+25=0
How do you solve this problem? 2x+9(root x) -5=0?
2x+9√(x)-5 = 0 (original equation) 2x+9√(x) = 5 (add 5) 9√(x) = 5-2x (subtract 2x) [9√(x)]² = (5-2x)² (square both sides) 81x = 25+4x²-20x (rewrite equation/expand right equation) 0 = 4x²-101x+25 (subtract 81x) 0 = (4x-1)(x-25) (factor equation) 4x-1 = 0 (original equation) 4x = 1 (add 1) x = 1/4 (divide by 4) x-25 = 0 (original equation) x = 25 (add 25) Check: 2(1/4)+9√(1/4)-5 = 0 1/2+9(1/2)-5 = 0 1/2+9/2-5 = 0 0 = 0 True - the solution checks correctly 2(25)+9√(25)-5 = 0 50+9(5)-5 = 0 50+45-5 = 0 90 = 0 False - the solution does not check correctly Solution: x = 1/4
What is [math] \sqrt{x} +3^2 = 2x [/math]?
sqrt(x) = 2x - 9x = (2x - 9)^2 = 4x^2 - 36x + 814x^2 - 37x+81 = 0x = ((37)+/-sqrt(16 - 4(4)(81)))/2(4)x = (37+/-sqrt(-1680))/8x = (37+/-16 sqrt (5)i)/8