Q. ax2+5x+2=0In the given equation, a is a constant. If the equation has the solutions x=−2 and x=−21, what is the value of a ?
Use Factored Form: Let's use the fact that if x=−2 and x=−21 are solutions to the equation ax2+5x+2=0, then the equation can be factored as a(x+2)(x+21)=0.
Expand and Compare: We can expand the factored form to get the standard form of the quadratic equation. Let's do that:a(x+2)(x+21)=ax(x+21)+2a(x+21)=ax2+(21)ax+2ax+a=ax2+(25)a+2ax+a.
Set Up Equations: Now, we compare the coefficients from the expanded form with the original equation ax2+5x+2=0. We have:Coefficient of x2: a (from both equations)Coefficient of x: (25)a+2a (from the expanded form) and 5 (from the original equation)Constant term: a (from the expanded form) and 2 (from the original equation)
Solve for a: Let's set up the equations for the coefficients of x and the constant term:(25)a+2a=5a=2
Check Constant Term: Now, we solve for a. Combine like terms in the coefficient equation:(25)a+2a=(25+24)a=(29)a=5
Check Constant Term: Now, we solve for a. Combine like terms in the coefficient equation: (25)a+2a=(25+24)a=(29)a=5 Multiply both sides by 92 to solve for a: a=(5×92)a=910
Check Constant Term: Now, we solve for a. Combine like terms in the coefficient equation: (25)a+2a=(25+24)a=(29)a=5Multiply both sides by 92 to solve for a: a=(5⋅92)a=910Now, let's check if the constant term also matches with a=910: a=910 The constant term in the expanded form is also a, which should be equal to 2. So, 910=2, which is not true. This means there is a mistake in our previous steps.
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