B oth values, 8 and –2, are solutions to the original equation.
A quadratic with a term missing is called an incomplete quadratic (as long as the ax2 term isn’t missing).
Solve x^2 –16=0.
Factor.
Tocheck,x2–16=0
We can also solve the equation by using the Sri Dharacharya formula:
The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:
x=−b±b2−4ac−−−−−−−√2ax=2a−b±b2−4ac
Solvex2+3x–4=0
This quadratic happens to factor:
x2+3x–4=(x+4)(x–1)=0
. ..so I already know that the solutions are x = –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:
T hen, as expected, the solution is x = –4, x = 1.
Thus , factorisation is the key in solving quadratic equations .
How do you solve this quadratic equation by the factorization method: 3 (2x+5) ^2=147?
3 (2x+5) ^2=147
3[(2x)^2+2(2x)(5)+(5)^2]=147
3(4x^2+20x+25)=147
12x^2+60x+75=147
12x^2+60x+75-147=0
12x^2+60x-72=0
12(x^2+5x-6)=0
x^2+6x-x-6=0
x(x+6)-1(x+6)=0
(x-1)=0 or (x+6)=0
x=1 or x=-6
What is the nature of the quadratic equation 5×2−2x−3=0 ?
5x²-5x+3x-3=0
5x(x-1)+3(x-1)=0
(5x+3)(x-1)=0
5x+3=0
5x=-3
X=-3/5
X-1=0
X=1
How can the quadratic equation x2−3x−10=0 be solved?
The quadratic equation x² – 3x – 10 = 0 can be solved by any one of the following four methods:
(1.) Factoring
The quadratic expression in x can be easily factored as follows:
x² – 3x – 10 = (x + 2)(x – 5) = 0
Now, by the Zero Product Property, i.e., if “a” and “b” are real numbers, then ab = 0 if and only if a = 0 or b = 0; therefore,
x + 2 = 0 or x – 5 = 0
x + 2 – 2 = 0 – 2 or x – 5 + 5 = 0 + 5
x = –2 or x = 5
Therefore, the solution set of our given quadratic equation is {–2, 5}.
(2.) The Quadratic Formula
The Quadratic Formula which is used to solv
What is the solution of the quadratic equation x^2 -2x – 3 = 0?
you are looking for two numbers that add to -2 and multiply to give -3
so start with the the -3. only two ways of factorising that
either -1,3 whch adds to 2
or 1,-3 which adds to -2, so that’s our answer
(x+1)(x-3)=0 ie the roots are at x=-1 and x=3 and the minimum is halfway at x=2
now i’ve given you that solution try this one x^2–5x+6=0
then investigate all the possible integer solutions of x^2+ax+6=0 where a is an integer. ie all the factorisations of 6
then plot all those solutions and see how the minimum point relates to a.
investigations will help you understand and learn.
copying the answers you get here will get you a tick on your homework book.
How do you solve the following quadratic equation: 6x^2 + 5x – 4 = 0?
6x^2+5x-4=0
Comparing with ax^2+bx+c=0
x=[ -b+ -(b^2-4ac)^1/2] /2 a
a=6 b=5 c=–4
x=[ –5+ – (25 -4×6×–4)^1/2 ] /12
=[ -5+– (121)^1/2 ]/12
x1=( -5 +11)/12 =6/12 =1/2
x2=( -5-11)/12 = -16/12 = -4/3
Roots of the given Quadra eqation are 1/2 and –4/3
Can the quadratic formula be used to solve 4x^3+23x^2-2x=0?
THIS QUADRATIC EQUATION IS NOT FACTORABLE WITHOUT USING THE QUADRATIC FORMULA…so,
Use the quadratic formula. The ‘a,’ in the formula is basically the coefficient of x^2, while the ‘b’, and ‘c’ values in the formula are basically the coefficients of x and the number without the variable at the end (the number before the 0, in this case, it is 2 – ‘c’ value).
After using the formula, the number will not be a perfect whole number. It would be (this is a general format – your answer may not completely resemble this): [a number +/- (a square root)] / another number.
P.S if your teacher hasn’t taught y
Using the quadratic formula, what are the solutions to the equation: 3x^2+2x+4=0?
You don’t need the quadratic formula for this because a quick inspection gives the answer x=3 and x=0.
However, if you want QF, it goes (3+or-sqrt(9–4*1*0))/2 so (3+or-sqrt(9))/2.
(3+or-3)/2 is 6/2 or 0/2, so answers of 3 and 0. I think 0/2 might be regarded as ‘undefined’, but 0 is a valid answer.
2x^2+3x+1=0 solve using a quadratic formula?
x={-1, -1/2}
PREMISES
2x^2+3x+1=0
CALCULATIONS
2x^2+3x+1=0
By factoring,
2x^2+3x+1=0 becomes
(2x+1)(x+1)=0
and,
Either 2x+1=0 and x=-1/2 and/or x+1=0 and x=-1
x=
{-1, -1/2}
By the “quadratic formula”, which can be used to solve any quadratic polynomial of the standard form ax^2 +/- bx +/- c, where a, b, and c are coefficients:
Step-by-step explanation:
This is the solution
Please Subscribe my yt channel JK Tutorial for more help.
Answer:
Solve x2–6x=16.
Followingthesteps,
x2–6x=16 becomes x2–6x–16=0
Factorisinggives:
(x–8)(x+2)=0
Settingeachfactortozero,
Thentocheck,
B oth values, 8 and –2, are solutions to the original equation.
A quadratic with a term missing is called an incomplete quadratic (as long as the ax2 term isn’t missing).
Solve x^2 –16=0.
Factor.
Tocheck,x2–16=0
We can also solve the equation by using the Sri Dharacharya formula:
The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:
x=−b±b2−4ac−−−−−−−√2ax=2a−b±b2−4ac
Solvex2+3x–4=0
This quadratic happens to factor:
x2+3x–4=(x+4)(x–1)=0
. ..so I already know that the solutions are x = –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:
x=−(3)±(3)2−4(1)(−4)−−−−−−−−−−−−−√2(1)x=2(1)−(3)±(3)2−4(1)(−4)
=−3±9+16−−−−−−√2=−3±25−−√2=2−3±9+16=2−3±25
=−3±52=−3−52,−3+52=2−3±5=2−3−5,2−3+5
=−82,22=−4,1=2−8,22=−4,1
T hen, as expected, the solution is x = –4, x = 1.
Thus , factorisation is the key in solving quadratic equations .
How do you solve this quadratic equation by the factorization method: 3 (2x+5) ^2=147?
3 (2x+5) ^2=147
3[(2x)^2+2(2x)(5)+(5)^2]=147
3(4x^2+20x+25)=147
12x^2+60x+75=147
12x^2+60x+75-147=0
12x^2+60x-72=0
12(x^2+5x-6)=0
x^2+6x-x-6=0
x(x+6)-1(x+6)=0
(x-1)=0 or (x+6)=0
x=1 or x=-6
What is the nature of the quadratic equation 5×2−2x−3=0 ?
5x²-5x+3x-3=0
5x(x-1)+3(x-1)=0
(5x+3)(x-1)=0
5x+3=0
5x=-3
X=-3/5
X-1=0
X=1
How can the quadratic equation x2−3x−10=0 be solved?
The quadratic equation x² – 3x – 10 = 0 can be solved by any one of the following four methods:
(1.) Factoring
The quadratic expression in x can be easily factored as follows:
x² – 3x – 10 = (x + 2)(x – 5) = 0
Now, by the Zero Product Property, i.e., if “a” and “b” are real numbers, then ab = 0 if and only if a = 0 or b = 0; therefore,
x + 2 = 0 or x – 5 = 0
x + 2 – 2 = 0 – 2 or x – 5 + 5 = 0 + 5
x = –2 or x = 5
Therefore, the solution set of our given quadratic equation is {–2, 5}.
(2.) The Quadratic Formula
The Quadratic Formula which is used to solv
What is the solution of the quadratic equation x^2 -2x – 3 = 0?
you are looking for two numbers that add to -2 and multiply to give -3
so start with the the -3. only two ways of factorising that
either -1,3 whch adds to 2
or 1,-3 which adds to -2, so that’s our answer
(x+1)(x-3)=0 ie the roots are at x=-1 and x=3 and the minimum is halfway at x=2
now i’ve given you that solution try this one x^2–5x+6=0
then investigate all the possible integer solutions of x^2+ax+6=0 where a is an integer. ie all the factorisations of 6
then plot all those solutions and see how the minimum point relates to a.
investigations will help you understand and learn.
copying the answers you get here will get you a tick on your homework book.
How do you solve the following quadratic equation: 6x^2 + 5x – 4 = 0?
6x^2+5x-4=0
Comparing with ax^2+bx+c=0
x=[ -b+ -(b^2-4ac)^1/2] /2 a
a=6 b=5 c=–4
x=[ –5+ – (25 -4×6×–4)^1/2 ] /12
=[ -5+– (121)^1/2 ]/12
x1=( -5 +11)/12 =6/12 =1/2
x2=( -5-11)/12 = -16/12 = -4/3
Roots of the given Quadra eqation are 1/2 and –4/3
Can the quadratic formula be used to solve 4x^3+23x^2-2x=0?
THIS QUADRATIC EQUATION IS NOT FACTORABLE WITHOUT USING THE QUADRATIC FORMULA…so,
Use the quadratic formula. The ‘a,’ in the formula is basically the coefficient of x^2, while the ‘b’, and ‘c’ values in the formula are basically the coefficients of x and the number without the variable at the end (the number before the 0, in this case, it is 2 – ‘c’ value).
After using the formula, the number will not be a perfect whole number. It would be (this is a general format – your answer may not completely resemble this): [a number +/- (a square root)] / another number.
P.S if your teacher hasn’t taught y
Using the quadratic formula, what are the solutions to the equation: 3x^2+2x+4=0?
3x^2 + 2x + 4 = 0
x = [- 2 (+/-) sqrt(4 – 4*3*4)]/6 = [- 2 (+/-) sqrt(-44)]/6 = [- 2 (+/-) 2 sqrt(-11)]/6 = [- 1 (+/-) i sqrt(11)]/3. Therefore:
x1 = -1/3 + i sqrt(11) and x2 = -1/3 – i sqrt(11)
X^2-3x=0 solve using a quadratic formula?
X^2-3x=0 solve using a quadratic formula?
You don’t need the quadratic formula for this because a quick inspection gives the answer x=3 and x=0.
However, if you want QF, it goes (3+or-sqrt(9–4*1*0))/2 so (3+or-sqrt(9))/2.
(3+or-3)/2 is 6/2 or 0/2, so answers of 3 and 0. I think 0/2 might be regarded as ‘undefined’, but 0 is a valid answer.
2x^2+3x+1=0 solve using a quadratic formula?
x={-1, -1/2}
PREMISES
2x^2+3x+1=0
CALCULATIONS
2x^2+3x+1=0
By factoring,
2x^2+3x+1=0 becomes
(2x+1)(x+1)=0
and,
Either 2x+1=0 and x=-1/2 and/or x+1=0 and x=-1
x=
{-1, -1/2}
By the “quadratic formula”, which can be used to solve any quadratic polynomial of the standard form ax^2 +/- bx +/- c, where a, b, and c are coefficients:
2x^2+3x+1=0 returns
x=[-3 +/-√(3^2-{4×2×1})]/(2×2)
x=[-3 +/- √(9-{8})]/4
x=[-3 +/- √1]/4
x=(-3 +/- 1)/4
x=
{-1, -1/2}
C.H.