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

Question

Pohanginah · Accepted Answer

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:   x2−3x+4x−12=0x2+x−12=0   Thus, x2+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:   x2−9x−x+9=0x2−10x+9=0   Thus, x2−10x+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:   x2+2x+8x+16=0x2+10x+16=0   Thus, x210x+16=0, is the standard form of the quadratic equation.

Nick Camelot · Accepted Answer

Step 1. Roots: 3, −4The quadratic equation in standard form is obtained by setting the roots as solutions and using the fact that for a quadratic equation ax2+bx+c=0, the roots satisfy the equation x=−b±b2−4ac2a. Let&#039;s substitute the roots into this equation:Let x1=3 and x2=−4. We have:x1=−b+b2−4ac2a3=−b+b2−4ac2a3=−b+b2−4ac2aSimilarly, substituting x2=−4 into the equation, we get:x2=−b−b2−4ac2a−4=−b−b2−4ac2aFrom 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, −2Following the same process as above, we substitute x1=−8 and x2=−2 into the quadratic equation formula:x1=−b+b2−4ac2a−8=−b+b2−4ac2ax2=−b−b2−4ac2a−2=−b−b2−4ac2aSolving these equations will give us the values of a, b, and c, which define the quadratic equation.Step 3. Roots: 1, 9Similar to the previous cases, we substitute x1=1 and x2=9 into the quadratic equation formula:x1=−b+b2−4ac2a1=−b+b2−4ac2ax2=−b−b2−4ac2a9=−b−b2−4ac2aSolving these equations will give us the values of a, b, and c, which define the quadratic equation.Step 4. Root: −5In this case, we have only one root, so we can directly write the equation as (x−(−5))2=0. Simplifying, we get x2+10x+25=0.Step 5. Roots: 10, 7Following the same process as above, we substitute x1=10 and x2=7 into the quadratic equation formula:x1=−b+b2−4ac2a10=−b+b2−4ac2ax2=−b−b2−4ac2a7=−b−b2−4ac2aSolving these equations will give us the values of a, b, and c, which define the quadratic equation.Step 6. Roots: −2, 15Similar to the previous cases, we substitute x1=−2 and x2=15 into the quadratic equation formula:x1=−b+b2−4ac2a−2=−b+b2−4ac2ax2=−b−b2−4ac2a15=−b−b2−4ac2aSolving these equations will give us the values of a, b, and c, which define the quadratic equation.

Mr Solver · Accepted Answer

Roots: 3,−4The quadratic equation is given by x2−(3+(−4))x+(3·−4)=0, which simplifies to x2−x−12=0.Roots: −8,−2The quadratic equation is given by x2−(−8+(−2))x+(−8·−2)=0, which simplifies to x2+10x−16=0.Roots: 1,9The quadratic equation is given by x2−(1+9)x+(1·9)=0, which simplifies to x2−10x+9=0.Root: −5The quadratic equation is given by (x−(−5))2=0, which simplifies to x2+10x+25=0.Roots: 10,7The quadratic equation is given by (x−10)(x−7)=0, which expands to x2−17x+70=0.Roots: −2,15The quadratic equation is given by (x−(−2))(x−15)=0, which expands to x2+17x−30=0.

madeleinejames20 · Accepted Answer

Solution:1. Roots: 3, -4To find the quadratic equation with these roots, we start by using the fact that the roots are solutions to the equation. Therefore, we have:(x−3)(x+4)=0Expanding this equation, we get:x2+x−12=0So, the quadratic equation with roots 3 and -4 in standard form is:x2+x−12=02. Roots: -8, -2Using the same approach, we can write the equation with these roots as:(x+8)(x+2)=0Expanding this equation, we get:x2+10x+16=0Thus, the quadratic equation with roots -8 and -2 in standard form is:x2+10x+16=03. Roots: 1, 9Applying the same method, we have:(x−1)(x−9)=0Expanding this equation, we get:x2−10x+9=0Hence, the quadratic equation with roots 1 and 9 in standard form is:x2−10x+9=04. Root: -5Since we have only one root, the quadratic equation can be written as:(x+5)=0Expanding this equation, we get:x+5=0Therefore, the quadratic equation with the root -5 in standard form is:x+5=05. Roots: 10, 7Applying the same approach, we can write the equation with these roots as:(x−10)(x−7)=0Expanding this equation, we get:x2−17x+70=0Thus, the quadratic equation with roots 10 and 7 in standard form is:x2−17x+70=06. Roots: -2, 15Using the same method, we have:(x+2)(x−15)=0Expanding this equation, we get:x2−13x−30=0Hence, the quadratic equation with roots -2 and 15 in standard form is:x2−13x−30=0

Write a quadratic equation in standard form with the given

Answered question

Answer & Explanation

New Questions in Algebra I