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Please help me with my math!!

Please help me with my math!!-example-1

2 Answers

2 votes

Answer:


  • {x}^(2) - 11x = - 30

Explanation:

To find:-

  • The equation which has roots 5 and 6 .

Answer:-

We are given that the roots of a quadratic equation are 5 and 6 , and we are interested in finding out the quadratic equation. For that as we know that , if
\alpha and
\beta are the roots of a quadratic equation, then ; the quadratic equation can be expressed as ,


\longrightarrow \boxed{x^2-(\alpha+\beta)x+(\alpha\beta) = 0 } \\

So , here the sum of the roots will be ,


  • \longrightarrow \alpha + \beta = 5+6=\boxed{11 } \\

And the product of roots , would be ,


  • \longrightarrow \alpha\beta = 6(5) = \boxed{30}\\

Now substitute the respective values, in the above mentioned formula as ,


\longrightarrow x^2 - (11)x + 30 = 0\\


\longrightarrow x^2-11x + 30 = 0\\

Subtract 30 on both the sides,


\longrightarrow \boxed{\boldsymbol{ x^2-11x = -30}} \\

This is the required quadratic equation .

User Kaykun
by
8.2k points
2 votes

Answer:


x^2-11x=-30

Explanation:

The roots of a quadratic equation are the x-values that make the quadratic equation equal to zero. These are also called the x-intercepts.

To determine the quadratic equation that has roots at x = 5 and x = 6, we can use the Intercept formula.


\boxed{\begin{minipage}{6.5 cm}\underline{Intercept form of a quadratic equation}\\\\$y=a(x-p)(x-q)$\\\\where:\\ \phantom{ww}$\bullet$ $p$ and $q$ are the $x$-intercepts. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}

Given the roots are x = 5 and x = 6, then:

  • p = 5
  • q = 6
  • y = 0

As the leading coefficient in all the answer options is 1, a = 1.

Substitute the values into the formula:


\implies 1(x-5)(x-6)=0


\implies (x-5)(x-6)=0

Expand the left side of the equation using the FOIL method:


\implies x^2-6x-5x+30=0


\implies x^2-11x+30=0

Subtract 30 from both sides of the equation:


\implies x^2-11x+30-30=0-30


\implies x^2-11x=-30

User Sulman
by
8.3k points

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