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# The integeral values of x for which x2 + 7x +13 is perfect square are _____ &_____

Arun
25763 Points
one year ago
Let \$f(x)=x^2+7x+13\$
\$x^2+7x+13 \$ is right between \$(x+3)^2, x^2+6x+9\$ and \$(x+4)^2,x^2+8x+16\$ for any integral value of \$x.\$
Now, subtract \$f(x)\$ from the nearest square higher than \$f(x)\$ , it will be equal to \$0 \$ .(*Remember we're finding perfect squares.)*
Here, \$(x+4)^2- f(x)=0\$
\$\implies x+3=0 \implies x=-3\$
Now, subtract the nearest square lower than \$f(x)\$ from \$f(x)\$ , it will be too , equal to zero.
Here , \$f(x)-(x+3)^2=0\$
\$\implies x+4=0 \implies x=-4\$
So the two values of \$x\$ are \$\boxed {-3,-4}.\$