# What are the coefficients of the quadratic equation ax^2+ bx + c =0 if it has the roots -4 and 13/7?

Asked on by michdon19

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

let f(x) = ax^2 + bx + c

given -4  and  13/7  are the roots.

Then (x+4) and (x- 13/7) are factors for f(x):

==> f(x) = (x+4)(x- 13/7)

= x^2 + 4x - 13/7 x  -52/7 = 0

= x^2  + 15/7 x -   52/7 = 0

Multiply by 7:

==> 7x^2 + 15x - 52 = 0

==> a= 7    b = 15     c = -52

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

roots  α = -4  and ß =13/7

x² -(α+ß)x + αß = 0

x² -(13/7-4)x - 4*13/7 = 0

x² -(-15/7)x - 52/7 = 0

x² + (15/7)x - 52/7 = 0

Multiply by 7 :

7x² + 15x - 52 = 0

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

As the quadratic equation has roots -4 and 13/7 we can use Vieta’s formula to find the required values. According to Vieta’s formula for a second degree equation, the sum of the roots of the equation ax^2 + bx + c = 0 is equal to –b/a and the product of the roots is c/a.

Hence we get:

-4+ 13/7 = -15/7 = -b/a

And (-4)* 13/7 =  52/7  = c/a

Therefore using these relations we get a = 7, b=15 and c =-52.

Therefore the quadratic equation we get is:

7x^2 + 5b - 52 = 0

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The given quadratic equation is ax^2+bx+c =0.

Given that this has roots -4 and 13/7.

We know that if x1 and x2 are the solutions of the quadratic equeion ax^2+bx+c = o, then ax^2+nx+c = a(x-x1)(x-x2) is an identiy.

So ax^2+bx+c = a(x-(-4))(x-13/7) is an identity.

ax^2+bx+c = ax^2 -a(-4+13/7)x -a*4*13/7

ax^2+bx+c = ax^2 +(15/7)ax-52a/7

Since the above is an identity, we can equate like terms on both sides:

b = 15a/7 and c = -52a/7.

Therefore for any a we choose  given b = 15a/y.  c = -52a/7.  But cannot be zero. If a = 0, the equation degenerates into a linear equation.

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