Answer:
We are given that the sum of their squares is equal to 313. This is a quadratic equation in a. We know that x=b±√b2−4ac2awhere ax2+bx+c=0 is a quadratic equation in x . Hence, the required consecutive positive integers are 12 and 13
Answer: 12 and 13
Step-by-step explanation:
Let the smaller number be a
,then the second number will be a+1
. Square these and add them and then equate the sum to 313. Then solve for a
.
Complete step-by-step answer:
. Now since the numbers are consecutive, the second number will be a+1
We are given that the sum of their squares is equal to 313.
a2+(a+1)2=313
…(1)
We know that (x+y)2=x2+y2+2xy
. We will use this to simplify (a+1)2
(a+1)2=a2+1+2a
We will substitute this in equation (1)
a2+a2+1+2a=313
2a2+2a+1=313
2a2+2a−312=0
Divide this by 2, we will get the following:
a2+a−156=0
This is a quadratic equation in a
We know that x=b±b2−4ac−−−−−−−√2a
where ax2+bx+c=0
is a quadratic equation in x
Using this, we get the following:
a=−1±(−1)2−4×1×(−156)−−−−−−−−−−−−−−−−−−−√2×1
a=−1±1+624−−−−−−√2
a=−1±625−−−√2
a=−1±252
a=−262,242
a=−13,12
Since, in the question we need positive integers so we will select a=12
So a+1=13
Hence, the required consecutive positive integers are 12
and 13
Note: You can check whether your answer is correct or not by substituting a=12
and a+1=13
in the given equation a2+(a+1)2=313
We get the following:
122+132=144+169=313
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Verified answer
Answer:
We are given that the sum of their squares is equal to 313. This is a quadratic equation in a. We know that x=b±√b2−4ac2awhere ax2+bx+c=0 is a quadratic equation in x . Hence, the required consecutive positive integers are 12 and 13
Answer: 12 and 13
Step-by-step explanation:
Let the smaller number be a
,then the second number will be a+1
. Square these and add them and then equate the sum to 313. Then solve for a
.
Complete step-by-step answer:
Let the smaller number be a
. Now since the numbers are consecutive, the second number will be a+1
.
We are given that the sum of their squares is equal to 313.
a2+(a+1)2=313
…(1)
We know that (x+y)2=x2+y2+2xy
. We will use this to simplify (a+1)2
(a+1)2=a2+1+2a
We will substitute this in equation (1)
a2+a2+1+2a=313
2a2+2a+1=313
2a2+2a−312=0
Divide this by 2, we will get the following:
a2+a−156=0
This is a quadratic equation in a
.
We know that x=b±b2−4ac−−−−−−−√2a
where ax2+bx+c=0
is a quadratic equation in x
.
Using this, we get the following:
a=−1±(−1)2−4×1×(−156)−−−−−−−−−−−−−−−−−−−√2×1
a=−1±1+624−−−−−−√2
a=−1±625−−−√2
a=−1±252
a=−262,242
a=−13,12
Since, in the question we need positive integers so we will select a=12
So a+1=13
Hence, the required consecutive positive integers are 12
and 13
Note: You can check whether your answer is correct or not by substituting a=12
and a+1=13
in the given equation a2+(a+1)2=313
We get the following:
122+132=144+169=313