Help it’s due tomorrow !!!!
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Answer:
9. x =0 or x = -3/2
Step-by-step explanation:
9.
4x^2 +6x =0. The first step is to factor x; x(4x+6) =0. This implies that either; x=0 or 4x+6 =0. Solving for x yields; 4x =-6 which upon dividing both sides by 4 becomes x =-3/2.
10.
7x^2 =21x. The equation can be rewritten as; 7x^2 -21x =0. We note that 7x is a common multiple and we factor it out; 7x(x-3)=0. This implies that either 7x =0 or x-3 =0. Solving 7x =0 yields; x=0. Solving x-3 = 0 yields; x =3. The solutions to the quadratic equation are thus; x =0 or x =3.
11.
(x+2)^2 =49. The equation is already in factored form. The next step is to obtain square roots on both sides of the equation which yields; (x+2) =±7. This implies that; x = -2±7. The solutions to the quadratic equation are thus; x = -9 or x = 5.
12.
x+3 =24x^2. The first step is to write the equation in the standard form; 24x^2 -x -3 =0. The next step we make the coefficient of x^2 equal to 1 by diving all through by 24; x^2 -(1/24)x - (1/8) =0. Consequently, we determine two numbers whose sum is -(1/24) and their product -(1/8). By trial and error the two numbers are found to be; -(3/8) and (1/3). The equation is then re-written as; x^2 +(1/3)x -(3/8)x -(1/8) =0. The equation is then factored as; x(x +1/3) -3/8 (x +1/3) =0. Upon simplification this becomes; (x -3/8)(x +1/3) =0. The solutions to the quadratic equation are thus; x =3/8 or x = -1/3.
13.
We employ the Desmos graphing utility to plot the functions and then determine the x-intercept which represents the zeros of the given function;
The graphical solutions to the first equation are; x =2.5 or x = -2.
14.
The graphical solutions to the this equation are; x =3.5 or x = -1.33.
15.
The graphical solutions to the this equation are; x =0.42 or x = -7.17.
16.
The graphical solutions to the this equation are; x =3.25 or x = -2.
17.
The graphical solutions to the this equation are; x =4.5 or x = -0.67.
18.
The graphical solutions to the this equation are; x =0.46 or x = -2.71.
19.
If 4 and -5 are the solutions to a quadratic equation, then; x =4 or x = -5. This implies that x -4 =0 or x +5 =0. Consequently, (x -4)(x+5) =0. Expanding the brackets and simplifying yields; x^2 +x -20 =0.
20.
If -6 and 0 are the solutions to a quadratic equation, then; x =-6 or x = 0. This implies that x +6 =0 or x =0. Consequently, x(x+ 6) =0. Expanding the brackets and simplifying yields; x^2 +6x =0.
21.
If 3 and 8 are the solutions to a quadratic equation, then; x =3 or x = 8. This implies that x -3 =0 or x -8 =0. Consequently, (x -3)(x -8) =0. Expanding the brackets and simplifying yields; x^2 -11x +24 =0