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A quadratic equation has a form:
`y=ax^2 + bx + c`
To determine the values of a, b and c, plug-in the given roots. Note that roots are the values of x when y is zero.
So, when we plug-in the root x=3, the resulting equation is:
`0=9a+3b+c` (Let this be EQ1.)
And, when we plug-in the roots x=-5, the resulting equation is:
`0=25a-5b+c` (Let this be EQ2.)
Next, plug-in the point (1, -12).
`-12=a+b+c` (Let this be EQ3.)
Now that we have three equations, to solve for values of a, b and c, apply elimination method.
To do so, subtract EQ3 from EQ2 to eliminate the c.
`6=4a+b` (Let this be EQ4.)
Next, subtract EQ3 from EQ2 to eliminate c again.
`2=4a-b` (Let this be EQ5.)
Then, add EQ4 and EQ5 to eliminate b.
Plug-in this value of a to either EQ4 or EQ5 to solve for b.
So, using EQ4, the value of b is:
Then, plug-in values of a and b to either EQ1, EQ2 or EQ3.
Using EQ3, the value of c is:
Now that the values of a, b and c are known, substitute their values to the form of quadratic equation which is y =ax^2+bx+c.
Hence, the quadratic equation that has roots x=3 and x=-5 and passes the point (1,-12) is `y=x^2+2x-15` .
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