Question

The quadratic equation $ax^2+20x+c=0$ has exactly one solution. If $a+c=29$, and $a

Answer 1

a² + c² = 641

Explanation:

Given :-

• ax² + 20x + c has exactly one solution .
• a + c = 29 .

For exactly one Solution ,

• b² - 4ac = 0
• 20² - 4*a*c = 0
• 4ac = 400
• ac = 100

Also ,

• a + c = 29
• ( a + c)² = 29²
• a² + c² + 2ac = 841
• a² + c² + 2*100 = 841
• a²+ c² = 841 - 200
• a² + c² = 641

Answer 2

The quadratic equation is:

4x² + 20x + 25 = 0

so a = 4 and c = 25

How to find the value of a and c?

ax² + 20x + c=0

We know that we have a single solution, so the discriminant must be zero, then we can write:

20² - 4*a*c = 0

And we know that:

a + c = 29

We can write the second equation as:

a = 29 - c

Replace that in the first equation:

20² - 4*(29 - c)*c = 0

400 + 4c² - 116c = 0

Divide all of this by 4

(400 + 4c² - 116c)/4 = 0

100 + c² - 29c = 0

Solving this we will get:

`c = (29 \pm √(29^2 - 4*100) )/(2) \\\\c = (29 \pm 21 )/(2)`

Then c can be:

c = (29 + 21)/2 = 25

And then:

a = 29 - 25 = 4

The quadratic is:

4x² + 20x + 25 = 0