Reggie

2021-08-16

Write a quadratic equation in standard form with the given root(s). 1. 3, -4 2. -8,-2 3. 1,9 4. -5 5. 10,7 6. -2,15

Pohanginah

The roots of the quadratic equations are: 3 and -4
Hence, (x−3)and (x+4) are the factors of the quadratic equation.
Thus, the quadratic equation is given by:
(x−3)(x+4)=0
Solving it, we get:
${x}^{2}-3x+4x-12=0{x}^{2}+x-12=0$
Thus, ${x}^{2}+x-12=0$, is the standard form of the quadratic equation.
The roots of the quadratic equations are: -8 and -2
Hence, (x+8)and (x+2) are the factors of the quadratic equation.
The roots of the quadratic equations are: 1 and 9.
Hence, (x−1)and (x−9) are the factors of the quadratic equation.
Thus, the quadratic equation is given by:
(x−1)(x−9)=0
Solving it, we get:
${x}^{2}-9x-x+9=0{x}^{2}-10x+9=0$
Thus, ${x}^{2}-10x+9=0$, is the standard form of the quadratic equation.
Thus, the quadratic equation is given by:
(x+8)(x+2)=0
Solving it, we get:
${x}^{2}+2x+8x+16=0{x}^{2}+10x+16=0$
Thus, ${x}^{2}10x+16=0$, is the standard form of the quadratic equation.

Nick Camelot

Step 1. Roots: $3$, $-4$
The quadratic equation in standard form is obtained by setting the roots as solutions and using the fact that for a quadratic equation $a{x}^{2}+bx+c=0$, the roots satisfy the equation $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$. Let's substitute the roots into this equation:
Let ${x}_{1}=3$ and ${x}_{2}=-4$. We have:
$\begin{array}{cc}\hfill {x}_{1}& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 3& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 3& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
Similarly, substituting ${x}_{2}=-4$ into the equation, we get:
$\begin{array}{cc}\hfill {x}_{2}& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill -4& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
From these equations, we can solve for $a$, $b$, and $c$. However, since the equation is not completely specified, there are infinitely many possible quadratic equations with these roots.
Step 2. Roots: $-8$, $-2$
Following the same process as above, we substitute ${x}_{1}=-8$ and ${x}_{2}=-2$ into the quadratic equation formula:
$\begin{array}{cc}\hfill {x}_{1}& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill -8& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
$\begin{array}{cc}\hfill {x}_{2}& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill -2& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
Solving these equations will give us the values of $a$, $b$, and $c$, which define the quadratic equation.
Step 3. Roots: $1$, $9$
Similar to the previous cases, we substitute ${x}_{1}=1$ and ${x}_{2}=9$ into the quadratic equation formula:
$\begin{array}{cc}\hfill {x}_{1}& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 1& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
$\begin{array}{cc}\hfill {x}_{2}& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 9& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
Solving these equations will give us the values of $a$, $b$, and $c$, which define the quadratic equation.
Step 4. Root: $-5$
In this case, we have only one root, so we can directly write the equation as $\left(x-\left(-5\right){\right)}^{2}=0$. Simplifying, we get ${x}^{2}+10x+25=0$.
Step 5. Roots: $10$, $7$
Following the same process as above, we substitute ${x}_{1}=10$ and ${x}_{2}=7$ into the quadratic equation formula:
$\begin{array}{cc}\hfill {x}_{1}& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 10& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
$\begin{array}{cc}\hfill {x}_{2}& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 7& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
Solving these equations will give us the values of $a$, $b$, and $c$, which define the quadratic equation.
Step 6. Roots: $-2$, $15$
Similar to the previous cases, we substitute ${x}_{1}=-2$ and ${x}_{2}=15$ into the quadratic equation formula:
$\begin{array}{cc}\hfill {x}_{1}& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill -2& =\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
$\begin{array}{cc}\hfill {x}_{2}& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \hfill 15& =\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
Solving these equations will give us the values of $a$, $b$, and $c$, which define the quadratic equation.

Mr Solver

Roots: $3,-4$
The quadratic equation is given by ${x}^{2}-\left(3+\left(-4\right)\right)x+\left(3·-4\right)=0$, which simplifies to ${x}^{2}-x-12=0$.
Roots: $-8,-2$
The quadratic equation is given by ${x}^{2}-\left(-8+\left(-2\right)\right)x+\left(-8·-2\right)=0$, which simplifies to ${x}^{2}+10x-16=0$.
Roots: $1,9$
The quadratic equation is given by ${x}^{2}-\left(1+9\right)x+\left(1·9\right)=0$, which simplifies to ${x}^{2}-10x+9=0$.
Root: $-5$
The quadratic equation is given by $\left(x-\left(-5\right){\right)}^{2}=0$, which simplifies to ${x}^{2}+10x+25=0$.
Roots: $10,7$
The quadratic equation is given by $\left(x-10\right)\left(x-7\right)=0$, which expands to ${x}^{2}-17x+70=0$.
Roots: $-2,15$
The quadratic equation is given by $\left(x-\left(-2\right)\right)\left(x-15\right)=0$, which expands to ${x}^{2}+17x-30=0$.

Solution:
1. Roots: 3, -4
To find the quadratic equation with these roots, we start by using the fact that the roots are solutions to the equation. Therefore, we have:
$\left(x-3\right)\left(x+4\right)=0$
Expanding this equation, we get:
${x}^{2}+x-12=0$
So, the quadratic equation with roots 3 and -4 in standard form is:
$\overline{){x}^{2}+x-12=0}$
2. Roots: -8, -2
Using the same approach, we can write the equation with these roots as:
$\left(x+8\right)\left(x+2\right)=0$
Expanding this equation, we get:
${x}^{2}+10x+16=0$
Thus, the quadratic equation with roots -8 and -2 in standard form is:
$\overline{){x}^{2}+10x+16=0}$
3. Roots: 1, 9
Applying the same method, we have:
$\left(x-1\right)\left(x-9\right)=0$
Expanding this equation, we get:
${x}^{2}-10x+9=0$
Hence, the quadratic equation with roots 1 and 9 in standard form is:
$\overline{){x}^{2}-10x+9=0}$
4. Root: -5
Since we have only one root, the quadratic equation can be written as:
$\left(x+5\right)=0$
Expanding this equation, we get:
$x+5=0$
Therefore, the quadratic equation with the root -5 in standard form is:
$\overline{)x+5=0}$
5. Roots: 10, 7
Applying the same approach, we can write the equation with these roots as:
$\left(x-10\right)\left(x-7\right)=0$
Expanding this equation, we get:
${x}^{2}-17x+70=0$
Thus, the quadratic equation with roots 10 and 7 in standard form is:
$\overline{){x}^{2}-17x+70=0}$
6. Roots: -2, 15
Using the same method, we have:
$\left(x+2\right)\left(x-15\right)=0$
Expanding this equation, we get:
${x}^{2}-13x-30=0$
Hence, the quadratic equation with roots -2 and 15 in standard form is:
$\overline{){x}^{2}-13x-30=0}$

Do you have a similar question?