Table of Contents
What are the solutions to the equation x2 6x 40?
The solutions to the equation x2 + 6x 40 are -10, 4.
What is the answer of x 2 6x 7 0?
The solutions to x2u22126xu221270 x 2 u2212 6 x u2212 7 0 are xu22121 and x7 .
What are the roots of the equation x2 6x 4 0?
The roots of the equation x2 + 6x 4 0 are u03b1, u03b2.
What are the roots of the quadratic equation x² 6x 7 0?
The roots of the equation x2u22126x+70 are u03b1 and u03b2. Find the equation with roots u03b1+1/u03b2 and u03b2+1/u03b1. Quadratic equation question, as specified in the title. The roots of the equation x2u22126x+70 are u03b1 and u03b2.
What are the solutions to the equation x2 6x?
The solutions to x2 – 6x 16 are x -2 and x 8.
Which is a solution to the equation?
A solution to an equation is a value of a variable that makes a true statement when substituted into the equation. The process of finding the solution to an equation is called solving the equation. To find the solution to an equation means to find the value of the variable that makes the equation true.
What is the solution of x2 6x 7 0?
The solutions to x2u22126xu221270 x 2 u2212 6 x u2212 7 0 are xu22121 and x7 .
How many solutions does x 2 6x 7 0 have?
Two solutions were found : x 7. x -1.
What is the roots of x² 6x 7 0?
The roots of the equation x2u22126x+70 are u03b1 and u03b2. Find the equation with roots u03b1+1/u03b2 and u03b2+1/u03b1. Quadratic equation question, as specified in the title. The roots of the equation x2u22126x+70 are u03b1 and u03b2.
How do you solve x 2 6x 7 by completing the square?
Step 1: Rearrange the equation in the form of ax2 + bx c, if necessary. Step 3: Factorize the sides using algebraic identity (a + b)2 into perfect squares. Step 4: Take square root on both the sides. Step 5: Solve for x.
What are the roots of the equation x2 6x 4?
The roots of the equation x2 + 6x 4 0 are u03b1, u03b2.
How do you find roots of an equation?
The roots are calculated using the formula, x (-b xb1 u221a (bxb2 – 4ac) )/2a. Discriminant is, D b2 – 4ac. If D x26gt; 0, then the equation has two real and distinct roots. If D x26lt; 0, the equation has two complex roots.
Which of the following is a quadratic equation x2 6x 4 0?
The roots of the equation x2u22126x+70 are u03b1 and u03b2. Find the equation with roots u03b1+1/u03b2 and u03b2+1/u03b1. Quadratic equation question, as specified in the title. The roots of the equation x2u22126x+70 are u03b1 and u03b2.
How do you find the roots of a quadratic equation?
The solutions to x2u22126xu221270 x 2 u2212 6 x u2212 7 0 are xu22121 and x7 .
How do you find the roots of a zero in a quadratic equation?
For a quadratic equation ax2 + bx + c 0,
- The roots are calculated using the formula, x (-b xb1 u221a (bxb2 – 4ac) )/2a.
- Discriminant is, D b2 – 4ac. If D x26gt; 0, then the equation has two real and distinct roots. If D x26lt; 0, the equation has two complex roots.
- Sum of the roots -b/a.
- Product of the roots c/a.
What is the shortcut to find the roots of a quadratic equation?
2 Answers
- 1) a+b+c0 then the roots are 1 and c/a. Example: 33x2u221241x+80 has 1 and 8/33 as solutions.
- 2) ba+c then the roots are u22121 and u2212c/a. Example: 1793×2+2016x+223 has u22121 and u2212223/1793 as roots.
- 3) for x2+bx+c0 I test the divisors of c. If one of them say u03b1 is a root then c/u03b1 is a root.
04-Jul-2016
What is the solution of x2 6x?
The solutions to x2 – 6x 16 are x -2 and x 8.
How do you factor x2 +6x?
The solutions to the equation x2 + 6x 40 are -10, 4.
