|Use the quadratic formula to solve this equation. d^2 + 7d -8 = 0
||Since the eqaution is in standard form, substitute values of a, b, and c into the quadratic fthe ormula. Thus, d= (-(1) +/- sqaure root ((7^2) – 4(1)(-8))/ 2(1). Multiply. Then, simplify the sqaure root to get the solutions d= -8 and 1.
|Use the quadratic formula to solve this equation. 1/2x^2 + 5x -1 = 0
||Clear the equation of fractions by multiplying both sides of the equation by the LCD 2. Substitute the values of a, b, and c into the quadratic formula and solve. The solutions for x are -5 +/- 3* the sq.rt. of 3.
|Solve. p^4 – 1296 = 0
||Factor the binomial into two binomials since this equation is the difference of two squares. Thus, the equation becomes (p^2 – 36)(p^2 + 36) = 0. Set each factor to 0 and solve for p. The solutions are 6, -6, 6i, -6i.
|Solve. x^2/3 – 2x^1/3 – 15 = 0
||Substitute the variable u in the original equation for x^1/3. The equation becomes u^2 – 2u – 15 = 0. Solve for u by factoring the equation. Thus, u = 5 or u = -3. Since x^2/3 = (x^1/3)^2, substitute x^1/3 back in for the variable u. x = 125, x = -27
|Solve. (2n + 1)^2 + 7(2n + 1) – 8 = 0
||Substitute variable y in where there is a (2n+1) since it is repeated. The equation becomes y^2+7y-8=0. Factor the equation (y+8)(y-1)=0. Set each factor to 0. Solve for y. y= -8 or 1. Sub in (2n+1) back for the variable y and solve for n. n= -9/2 and 0
|Find all real and complex roots. x^4 – 19x^2 + 48 = 0
||First, sub in a linear term for the variable raised to the lowest power. Let y=x^2; y^2=x^4. The new equation is y^2-19y+48=0. Factor. (y-16)(y-3)=0. Set factors to 0. y=16 or 3. Sub x^2 back for variable y to find the original value, x. x=+/-4; +/- sqrt3
|Solve. y^3 + 25y -3y^2 – 75 = 0
||Factor by grouping. Group so that each group has a common factor. (y^3+25y)-(3y^2-75)=0. Factor out the GCF from each group. y(y^2+25)-3(y^2+25)=0. Factor out common factor (y^2+25). (y-3)(y^2+25)=0. Set factors to 0. Solve for y. Solutions are y=3; +/-5i