Write a quadratic equation in standard form such that -5 is it's only solution.

[tex]\displaystyle\bf\\x_1=x_2=-5\\\\(x-x_1)(x-x_2)=0\\\\(x-(-5))(x-(-5))=0\\\\(x+5)(x+5)=0\\\\x^2+2\cdot x\cdot5 + 5^2=0\\\\\boxed{\bf x^2+10x + 25=0}[/tex]

Answer Link

yoodyannapolis yoodyannapolis · Answer 2 · 2021-11-03T11:25:01+01:00

The quadratic equation such that -5 is it's only solution is [tex]f(x)=x^{2} + 10x+25[/tex].

The standard quadratic equation is [tex]ax^{2} +bx+c=0[/tex]

Zero of the quadratic equation = -5

Also quadratic equation

[tex]f(x)=x^{2} - Sx+P[/tex]

Where S = sum of zeroes

P = product of zeroes

So, sum of zeroes = -5 + (-5)

= -10

Product of zeroes = -5×-5

=25

Now, [tex]f(x)=x^{2} + 10x+25[/tex]

Therefore quadratic equation in standard form such that -5 is it's only solution is [tex]f(x)=x^{2} + 10x+25[/tex].

Learn more:https://brainly.com/question/16546913