Question

Please help me with my math!!

Answer 1

• 
  `{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 .

Answer 2

`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`