How do you solve ${{x}^{2}}+6x-7=0$?
Answer
623.7k+ views
Hint: In this question, we are given a quadratic equation and we need to find the value of x which satisfies this equation. For this we will split the middle term method. For an equation of the form $a{{x}^{2}}+bx+c=0$ we need to find two numbers ${{n}_{1}},{{n}_{2}}$ to split the middle term (b) such that ${{n}_{1}}+{{n}_{2}}=b\text{ and }{{n}_{1}}\cdot {{n}_{2}}=ac$. After splitting we will take common terms out of the first two terms and common terms from the last two terms such that after taking common we are left with the same terms. This will help us in factoring. Then putting each factor equal to 0 we will get the value of x.
Complete step-by-step answer:
Here we are given the equation as ${{x}^{2}}+6x-7=0$. The equation is a quadratic equation (of degree 2) so we will get two values of x. Let us use the split middle term method for solving this equation. We have the equation as ${{x}^{2}}+6x-7=0$.
We know that for an equation of the form $a{{x}^{2}}+bx+c=0$ we need ${{n}_{1}},{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=b\text{ and }{{n}_{1}}\cdot {{n}_{2}}=ac$. So here we have a = 1, b = 6 and c = -7.
So we need ${{n}_{1}},{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=6\text{ and }{{n}_{1}}\cdot {{n}_{2}}=-7$.
As we know only two numbers which when multiplied gives 7 are 1 and 7. So if we take 7 and -1 we get 7-1 = 6 and (7)(1) = 7 that satisfies both conditions. So ${{n}_{1}}=7\text{ and }{{n}_{2}}=-1$.
Splitting the middle term in this way we get ${{x}^{2}}+\left( 7-1 \right)x-7=0\Rightarrow {{x}^{2}}+7x-x-7=0$.
Let us take x common from the first two terms and -1 from the last two terms so we have $x\left( x+7 \right)-1\left( x+7 \right)=0$.
As we can see after taking common terms we are left with (x+7) from both. So let us take (x+7) common from both terms we get $\left( x-1 \right)\left( x+7 \right)=0$.
Putting both these factors as equal to 0 we have,
(I) x-1 = 0
Adding 1 to both sides we get,
$x-1+1=0+1\Rightarrow x=1$.
(II) x+7 = 0.
Subtracting 7 from both sides we get,
\[x+7-7=0-7\Rightarrow x=-7\].
Hence the two values of x are 1 and -7 and these values will satisfy the given equation.
Note: Students should take care of the signs while selecting the number ${{n}_{1}},{{n}_{2}}$. Note that we must be left with the same factor (x+7) so that we can factorize the equation. Number of solutions is the same as that of the degree of the equation. We could also use a quadratic formula for finding the solution.
Complete step-by-step answer:
Here we are given the equation as ${{x}^{2}}+6x-7=0$. The equation is a quadratic equation (of degree 2) so we will get two values of x. Let us use the split middle term method for solving this equation. We have the equation as ${{x}^{2}}+6x-7=0$.
We know that for an equation of the form $a{{x}^{2}}+bx+c=0$ we need ${{n}_{1}},{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=b\text{ and }{{n}_{1}}\cdot {{n}_{2}}=ac$. So here we have a = 1, b = 6 and c = -7.
So we need ${{n}_{1}},{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=6\text{ and }{{n}_{1}}\cdot {{n}_{2}}=-7$.
As we know only two numbers which when multiplied gives 7 are 1 and 7. So if we take 7 and -1 we get 7-1 = 6 and (7)(1) = 7 that satisfies both conditions. So ${{n}_{1}}=7\text{ and }{{n}_{2}}=-1$.
Splitting the middle term in this way we get ${{x}^{2}}+\left( 7-1 \right)x-7=0\Rightarrow {{x}^{2}}+7x-x-7=0$.
Let us take x common from the first two terms and -1 from the last two terms so we have $x\left( x+7 \right)-1\left( x+7 \right)=0$.
As we can see after taking common terms we are left with (x+7) from both. So let us take (x+7) common from both terms we get $\left( x-1 \right)\left( x+7 \right)=0$.
Putting both these factors as equal to 0 we have,
(I) x-1 = 0
Adding 1 to both sides we get,
$x-1+1=0+1\Rightarrow x=1$.
(II) x+7 = 0.
Subtracting 7 from both sides we get,
\[x+7-7=0-7\Rightarrow x=-7\].
Hence the two values of x are 1 and -7 and these values will satisfy the given equation.
Note: Students should take care of the signs while selecting the number ${{n}_{1}},{{n}_{2}}$. Note that we must be left with the same factor (x+7) so that we can factorize the equation. Number of solutions is the same as that of the degree of the equation. We could also use a quadratic formula for finding the solution.
Recently Updated Pages
A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

Write any three differences between metals and nonmetals class 10 social science CBSE

Amit standing on a horizontal plane finds a bird flying class 10 maths CBSE

Two circles of radii 5 cm and 3 cm intersect at two class 10 maths CBSE

Solve the following i John and Jivanti together have class 10 maths CBSE

What is the relation between orthocenter circumcentre class 10 maths CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Which country is known as "The land of Fire and Ice"?

10 examples of evaporation in daily life with explanations

What is the full form of POSCO class 10 social science CBSE

Make a sketch of the human nerve cell What function class 10 biology CBSE

Differentiate between Xylem and phloem class 10 biology CBSE

