# If a quadratic equation ax^2 + 8x + 9 = 0 has two equal roots what is the value of a?

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

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ax^2 + 8x + 9 =0

==> a=a    b= 8     and c = 9

if the quadratic equation has one root, then we know that delta should be zero:

Delta = b^2 - 4ac

==> 0 = 8^2 - 4*a*9

==> 64 - 36a = 0

Subtract 64 from both sides:

==> -36a = -64

Divide by -36

==> a = -64/-36

==> a = 16/9

Then the values of a in order to have one root is a= 16/9

justaguide | College Teacher | (Level 2) Distinguished Educator

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The roots of a quadratic equation are given by the expressions [–b + sqrt (b^2 – 4ac)]/2a and [–b - sqrt (b^2 – 4ac)]/2a. Now in the question we have to find the value of a for which the quadratic equation has two equal roots. So we need to have:

[–b + sqrt (b^2 – 4ac)]/2a = [–b - sqrt (b^2 – 4ac)]/2a

=> sqrt (b^2 – 4ac) = 0

=> (b^2 – 4ac) =0

=> b^2 = 4ac

=> a = b^2 / 4c

as b = 8 and c = 9

=> a = 8^2 / 4*9

=> a = 16/9

Therefore to satisfy the given condition a should be equal to 16/9

neela | High School Teacher | (Level 3) Valedictorian

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ax^2+8x+9 =0

To find the the value of a , if ax^2+8x+9 has only one root. But no quadratic equation has only one one root. A quadratic equation has always two roots. So we presume that the given quadratic equation has two identical roots and proceed.

We know that a quadratic equation Ax^2+Bx+C = 0 has two roots given by  x1 = {-b+sqrt(b^2-4ac)}/2a and x 2 = {-b+sqrt(b^2-4ac)}/2.

Since this quadratic equation has only one root, x1 = x2.

Which is possible only when b^2-4ac = 0.

Therefore (b^2-4ac = ).

Therefore 8^2 - 4*9a=0

64 = 36a.

Or a = 64/36.

So if a = 64/36 = 16/9, then the given expression has only one root (or two identical roots).

Tally:

When a = 16/9  the equation becomes:

16/9x^2+8x+9 = 0 is a quadratic equation.

(8x/6)^2 +8x+3^2 = 0. Or

(8x/3+3)^2=0.

So the root is 8x/3 +3 = 0, Or x = -9/8 when a = 8/3.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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A quadratic equation has always 2 roots. When you say that the equation has "only one root", you are wrong. In fact, the equation has 2 equal roots.

ax^2 + 8x + 9 = 0

We'll use Viete's relations between coefficients and roots:

x1 + x2 = -8/a (1)

x1*x2 = 9/a (2)

But x1 = x2, because the roots are equal

x1 + x1 = -8/a

2x1 = -8/a

x1 = -4/a

x1^2 = 9/a => (-4/a)^2 = 9/a

We'll square raise and we'll subtract 9/a both sides:

16/a^2 - 9/a = 0

We'll eliminate the denominator:

16 - 9a = 0

9a = 16

We'll divide by 9 both sides:

a = 16/9

So, for the quadratic equatino to have 2 equal roots, the value of the coefficient a is 16/9.