Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

ax^(2)+5x+2=0
In the given equation, 
a is a constant. If the equation has the solutions 
x=-2 and 
x=-(1)/(2), what is the value of 
a ?

ax2+5x+2=0 a x^{2}+5 x+2=0 \newlineIn the given equation, a a is a constant. If the equation has the solutions x=2 x=-2 and x=12 x=-\frac{1}{2} , what is the value of a a ?

Full solution

Q. ax2+5x+2=0 a x^{2}+5 x+2=0 \newlineIn the given equation, a a is a constant. If the equation has the solutions x=2 x=-2 and x=12 x=-\frac{1}{2} , what is the value of a a ?
  1. Use Factored Form: Let's use the fact that if x=2x = -2 and x=12x = -\frac{1}{2} are solutions to the equation ax2+5x+2=0ax^2 + 5x + 2 = 0, then the equation can be factored as a(x+2)(x+12)=0a(x + 2)(x + \frac{1}{2}) = 0.
  2. Expand and Compare: We can expand the factored form to get the standard form of the quadratic equation. Let's do that:\newlinea(x+2)(x+12)=ax(x+12)+2a(x+12)=ax2+(12)ax+2ax+a=ax2+(52)a+2ax+aa(x + 2)(x + \frac{1}{2}) = ax(x + \frac{1}{2}) + 2a(x + \frac{1}{2}) = ax^2 + (\frac{1}{2})ax + 2ax + a = ax^2 + (\frac{5}{2})a + 2ax + a.
  3. Set Up Equations: Now, we compare the coefficients from the expanded form with the original equation ax2+5x+2=0ax^2 + 5x + 2 = 0. We have:\newlineCoefficient of x2x^2: aa (from both equations)\newlineCoefficient of xx: (52)a+2a(\frac{5}{2})a + 2a (from the expanded form) and 55 (from the original equation)\newlineConstant term: aa (from the expanded form) and 22 (from the original equation)
  4. Solve for aa: Let's set up the equations for the coefficients of xx and the constant term:\newline(52)a+2a=5(\frac{5}{2})a + 2a = 5\newlinea=2a = 2
  5. Check Constant Term: Now, we solve for aa. Combine like terms in the coefficient equation:\newline(52)a+2a=(52+42)a=(92)a=5(\frac{5}{2})a + 2a = (\frac{5}{2} + \frac{4}{2})a = (\frac{9}{2})a = 5
  6. Check Constant Term: Now, we solve for aa. Combine like terms in the coefficient equation: (52)a+2a=(52+42)a=(92)a=5(\frac{5}{2})a + 2a = (\frac{5}{2} + \frac{4}{2})a = (\frac{9}{2})a = 5 Multiply both sides by 29\frac{2}{9} to solve for aa: a=(5×29)a = (5 \times \frac{2}{9}) a=109a = \frac{10}{9}
  7. Check Constant Term: Now, we solve for aa. Combine like terms in the coefficient equation: (52)a+2a=(52+42)a=(92)a=5(\frac{5}{2})a + 2a = (\frac{5}{2} + \frac{4}{2})a = (\frac{9}{2})a = 5Multiply both sides by 29\frac{2}{9} to solve for aa: a=(529)a = (5 \cdot \frac{2}{9}) a=109a = \frac{10}{9}Now, let's check if the constant term also matches with a=109a = \frac{10}{9}: a=109a = \frac{10}{9} The constant term in the expanded form is also aa, which should be equal to 22. So, 109=2\frac{10}{9} = 2, which is not true. This means there is a mistake in our previous steps.

More problems from Write and solve direct variation equations